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Independent Project: Cirrus SR22 Welcome to Engadin Airport boom + . Zack Hendry Swiss HB-KHG . College of Aeronautics, Embry-Riddle Aeronautical University . ASCI 309: Aerodynamics . Captain Jack Vandelaar . Jan. 29, 2023 PHOTOS. NET This Photo by Unknown Author is licensed under CC BY-S2.3 Aircraft Selection, Airfoils Research, & Physical Laws N723CD Cirrus SR22 / Roncz Airfoil This Photo by Unknown Author is licensed under CC BY-NC-ND\fKTAS @ 150 kts Compressibility correction -30 EAS = CAS + AVc . Using 150 as the indicated M = 1.00 airspeed. -20 45 000 . Due to airspeed and altitude 60 000 being low, under 10,000 ft and Compressibility correction AVC, kts. Airspeed 4 55 000 40 000 35 000 30 000 under 200 knots, EAS is 60 000 25 090 20 000 equivalent to CAS. -10 - 15000 . No corrections are made. 10 000 Pressure . KTAS=- KEAS 150 5 000 = 150 altitude Vo VI Sea level ft. 100 200 300 400 500 600 Calibrated airspeed, kts.Formulas Applied - Wing Area = S = b (wingspan) * ca', (average chord) =38.5 * 3.764 = 144.9 - Aspect Ratio = AR = i av 38.5 = 3.764 - e= efficiency of wing = assumed value of .85 = 10.2 ' Dynamic Pressure = q = U mef / 295 = 1 (164.282) / 295 = 91.48 ' Coefficient of Lift = CL = W/qS = 3,000/ 1 (144.9): 1.34 2+3+ 4. 31467 22 Formulas Applied cont. + b' =X xx X C 2 ZT = C . Coefficient of Induced Drag = CDI= IT*e*AR 1.342 1.7956 = .0659 T[*.85*10.2 27.2376 . Function of the Coefficient of Parasite Drag Cop = .021 x 2+ . CD = Coefficient of Drag = Cop+ CDI 4= 14 ! 0 . Dp=Parasite Drag = CDpqS DX IN90 . X _ 1 .021*15.44*144.9 = 46.98 -2. 1964 . D =Induced Drag= CDIqS 066*15.44*144.9= 147.7 A . DT= Total Drag = Di+ DpWhat are the minimum drag parameters for your aircraft? . Minimum drag value D(min) CIRRUS = 166.6 lbs FLX - KID . Speed VD(min) at which minimum drag occurs = 89.88 Kts . Relationship between Do and D; at D(min) Do and D, are inverse of each other the two values intersect at D(min) NET This Photo by Unknown Author is licensed under CC BY-SWhat are the maximum lift to drag ratio % parameters D for your aircraft? CL ' c max value D = 18.01 0 Speed at which gL max occurs D =89.88 kts Glide Performance BM"): W / (L/D)ma,= 3000/ 18.01 = 166.6 Lift is proportional to distance and drag is proportional to altitude L/D is proportional to glide ratio (distance/altitude). Using the CL at Dtmln) of .76 the correlating AOA is 5.125" 5.125\" is therefore the best AOA for best glide conditions 5.2 Unaccelerated Performance This Photo by Unknown Author is licensed under CC BY-SCirrus SR22 Roncz Airfoil . Wingspan = 38.5 ft . Wing Area (S)=144.9ft2 CIRRUS. . Aspect Ratio = 10.2 . . L=Lift = Weight = W, 3,000 . CLMAY = Max coefficient of lift, 1.3408 . CDP= . Efficiency Factor= .85 . Weight=3,000 lbs. This Photo by Unknown Author is licensed under CC BY-NOPower-required P TVk DV k . Horsepower HP= 325 Pr = 325 Drag (lbs) V (KTAS)\fMaximum range airspeed . Maximum Range is the furthest distance an aircraft can fly . Maximum Range is achievable by cruising at the L/D MAX . L/D MAX = 18.01 . The Cirrus SR22 can achieve L/D MAX at approximately 90 knotsMaximum endurance airspeed Max Endurance is the airspeed at which an aircraft can stay aloft the longest amount of time This will be achievable at the minimum drag to the power required According to the power required graph, that value is = 46 HP at a speed of 90 knots Maximum forward airspeed . Max forward speed is where Pr=Pa . Based on the power required * * table . Using a maximum power available of 310 HP . The Cirrus SR22 will have a max airspeed of 211 knotsBest glide airspeed . At CL C max value of 18.01 the highest lift to drag ratio is produced. . The Cirrus SR22 should therefore travel 18.01 feet for every foot OLD of altitude. . At 10,000 feet AGL 180,100 ft or 29.64 nautical miles if traveling 90 knots at the time of engine failure This Photo by Unknown Author is licensed under CC BY-SBest angle of climb (AOC) & associated airspeed . Maximum climb angle AOC or achieving maximum height in the shortest distance is Cirrus calculated by the formula 325 PA -PR N830CD SR22 . AOC sin y = V. W 325 (310 -40.43) . AOC sin y = S 67.5 *3,000 25.63 . At 67.5 knots - - DAMMNAIAD TIT TL This Photo by Unknown Author is licensed under CC BY-NC-NDBest rate of climb (ROC) & associated airspeed . The best rate of climb (ROC) is the highest achievable altitude in the shortest amount of time N166RT . Found at L/Dmax or 90 knots . Calculated by ROC = 33,000 (Pa -Pr) W 33,000 (310 -46.14) 3,000 . = 2,902.46 fpm This Photo by Unknown Author is licensed under CC BY-NC-ND\fLighter Load = Less Power Req'd G) If weight were reduced from 3,000 lbs to an Maximum Range airspeed goes down to 80 empty weight of 2,250 lbs knots compared to the 90 knots at 3,000 lbs Cl v Alpha Stall AOA and 1.40 associated CLmax 1.20 1.00 0.80 . Stall AOA = 16.50 0.60 . CL Max = 1.3408 0.40 . -0.55 degrees = 0 Lift 0.20 0.00 -0.20 0.40 0.60 -10.0 -5.0 0.0 5.0 10.0 15.0 20.(\fTakeoff THIIIIIIIIIIIIIIII 6000 Takeoff speed Associated conditions Weight pounds Liftoff 50 ft Power Full throttle 2600 rpm MPH Guide lines not applicable for kts kts MPH 5000 Mixture Learn to appropriate 2950 66 76 72 83 Reference line fuel pressure Reference line Reference line intermediate 2800 64 74 70 81 Flaps Up 2600 63 72 68 78 Landing Retrack after positive 4000 Obstacle heights 2400 61 70 66 76 gear climb established callwing 2200 58 67 63 73 Cowl Open flaps 3000 . Takeoff speed for a selected Headwind takeoff weight: 67 kts for 10000. Pressure altitude - feet ISA 2000 2,500lbs 8000 . Length of the takeoff roll: 6080 1000 4000 2000 . 1,000 ft. C -40 -30 -20 -10 0 10 20 30 40 50 2800 2600 2400 2200 0 10 20 30 0 50 Weight Wind component Obstacle . The distance required to Outside air temperature (1b) (knots) height (ft) clear a 50 ft obstacle: F -40 -20 0 20 40 60 80 100 120 . 1,000 ft. Figure 8.1. Takeoff distance graph. Source: U.S. Department of Transportation Federal Aviation Administration (2008a).Takeoff #1 6,000 Takeoff speed Associated conditions Weight pounds Lift-off 50 f Power Full throttle 2,600 rpm Guide lines not applicable for kts MPH kts MPH 5,000 Mixture Learn to appropriate 2,950 76 Reference line 66 72 83 Reference line fuel pressure Reference line Intermediate 2.800 64 74 70 81 Flaps Up 2,600 63 72 68 78 Landing Retract after positive 4,000 Obstacle heights . Ex. 1 Yellow Line 2,400 61 70 66 76 gear climb established 2.200 58 67 63 73 | Cowl Open flaps 3,000 . Temp. 25 C . Weight 2,500 lbs. 10/0oo Pressure altitude - feet ISA 2,000 . Headwind 10 kts 8 000 6,060 1,000 ,000 . 50-foot obstacle 72.000 . Take off distance 1,000 ft. C -40 -30 -20 -10 0 10 20 30 40 50 2,800 2,600 2,400 2,200 0 10 20 30 50 Outside air temperature Weight Wind component Obstacle (pounds) (knots) height (feet) F -40 -20 0 20 40 60 80 100 120 Figure 8.1. Takeoff distance graph. Source: U.S. Department of Transportation Federal Aviation Administration (2008a).Takeoff #2 6,000 Takeoff speed Associated conditions Weight Lift-off 50 ft pounds Power Full throttle 2,600 rpm fide Ines not applicable for kts MPH kts MPH 5,000 Mixture Learn to appropriate 2,950 Reference line 56 76 72 Reference line fuel pressure Reference line Intermediate 2,800 64 74 70 81 Guide Flaps Up . Ex. 2 Green Line 2.600 63 72 68 78 Landing Retract after positive 4,000 2,400 61 70 66 76 gear climb established Obstacle 2,200 58 67 63 73 Cowl Open . Temp. 30 C flaps 3,000 . Weight 2,500 lbs 10/0oo Pressure altitude - feet ISA 2,000 . Headwind 15 kts 8 009 6,909 1,000 41060 . 50-foot obstacle . Take off distance 500 ft. C -40 -30 -20 -10 0 10 20 30 40 50 2,800 2,600 2,400 2,200 0 10 20 30 Outside air temperature Weight Wind component Obstacle (pounds) (knots) height (feet) F -40 -20 0 20 40 60 80 100 120 Figure 8.1. Takeoff distance graph. Source: U.S. Department of Transportation Federal Aviation Administration (2008a).Landing a Cirrus SR22 Approach speed for a selected landing weight: 90 kts @ 3,600 lbs . Drag during landing: 200 lbs. . Stall speed = 74 knots . Total drag at touchdown (stall speed) 233.4lbs. . Average drag during landing roll (based on total drag at stall speed 216.7lbs\fResidual thrust q = d x V^2 / CDI = V (KTAS) Dp = CDp CL = W/qs [1/ (reAR)] CD = Di = Dt = Di + Pr = HP = Pp = Pi = 295 (1b/ft^2) CDp CDP+CDI CL/CD q s Cdi q s Dp DtVk /325 DiVk CL^2 DiVk Power Excess (lb) (lb) (lb) /325 /325 2 Availble Power 74.0 18.56 1.34 0.066 0.021 0.09 15.43 6.48 176.9 233.4 . Residual power based on the 53.14 12.86 40.28 310 80 256.86 21.69 1.15 0.048 0.021 0.07 16.56 66.02 151.4 217.4 53.51 16.25 37.26 310 256.49 rated power of the aircraft 90 27.46 0.90 0.030 0.021 0.05 17.72 83.55 119.6 203.1 56.26 23.14 33.12 310 253.74 100 33.90 0.73 0.020 0.021 0.04 8.00 103.15 96.9 200.0 . Power required at 61.54 31.74 29.81 310 248.46 110 41.02 0.61 0.013 0.021 0.03 17.57 124.81 80.1 204.9 59.3 42.24 7.10 310 240.66 touchdown = 53.14 120 48.81 0.51 0.010 0.021 0.03 16.68 148.53 67.3 215.8 79.68 54.84 24.84 310 230.32 130 57.29 0.43 0.007 0.021 0.03 15.54 174.32 57.3 231.6 92.66 . Total power= 310 -53.14 69.73 22.93 310 217.34 140 66.44 0.37 0.005 0.021 0.03 14.31 202.17 49.4 251.6 108.38 87.09 21.29 310 201.62 150 76.27 0.33 0.004 . = Residual thrust of 256.86 0.021 0.02 13.08 232.09 43.1 275.1 126.99 107.12 19.87 310 183.01 160 86.78 0.29 0.003 0.021 0.02 11.92 264.06 37.8 301.9 148.63 lbs. 130.00 18.63 310 161.37 170 97.97 0.25 0.002 0.021 0.02 10.86 298.10 33.5 331.6 173.46 155.93 17.53 310 136.54 180 109.83 0.23 0.002 0.021 0.02 9.89 334.20 29.9 364.1 201.66 . Thrust at touchdown: 53.14 lbs. 185.10 16.56 310 108.34 190 122.37 0.20 0.002 0.021 0.02 9.02 372.37 26.8 399.2 233.38 217.69 5.69 310 76.62 200 135.59 0.18 0.001 0.021 0.02 8.24 412.60 24.2 436.8 268.81 . Average thrust during the 253.91 14.90 310 41.19 210 149.49 0.17 0.001 0.021 0.02 7.55 454.89 22.0 476.9 308.12 293.93 .4.19 310 1.88 landing roll 57.34 220 164.07 0.15 0.001 0.021 0.02 6.9 499.24 20.0 519.3 351.50 337.95 13.55 310 -41.50 230 179.32 0.14 0.001 0.021 0.02 6.38 545.66 18.3 564.0 399.12 386.16 12.96 310 -89.12Average deceleration during the landing roll To determine avg. deceleration, we need to figure mass, residual thrust, drag, and friction from nose and main landing gear M = WT / 32.2 M = 3,600 /32.2 = 111.8 slugs Friction front gear: F, = .02*.25*3600 = 18 Friction rear gear = Fb .73*.75*36OO =197l The formula for Deceleration is = T-D- F,- Fb /M A: 256.86-18-1971/ 118.8 2 -14.68 ft/S in l'lu (w w Landing Distance - With approach speed we can calculate landing distance Approach Speed = 74 knots - Knots to ft/s 74 * 1.69 = 125.06 S = V2 / 2A = Landing Distance - 125.062/ 2 (14.68) = 532.69 Influence of altitude on landings Altitude effects air density, pressure, and temperature in turn effecting density altitude Increasing density altitude will increase the landing speed Influence on landing speeds: - IAS will remain the same, but TAS will increase Influence on landing distances: The FAA's Airplane Flying Handbook (2016) states that a 10% increase in landing speed requires a 21% increase or more in landing distance. Comparison of Cirrus SR22 airfoil to a symmetric airfoil Cl v Alpha Cl v Alpha 1.40 1.00 1.20 0.80 1.00 0.60 0.80 0.40- 0.60 0.20 0.40 0.00- 0.20 -0.20 - 0.00 -0.40 - -0.20 -0.60- -0.40- 0.80 - Marske7-il (Cambered) -0.60 HT05-il (Symmetrical) -10.0 -5.0 10.0 1.00 5.0 10.0 15.0 20.( 0.15.0 -10.0 -5.0 0.0 5.0 10.0 15.( .16Effects of Altitude . Using the equation s S2 1 02 . S2 = Altitude landing distance . S, = Standard Sea Level landing distance . 02 = Altitude density ratio $2- _ S2 1 S1 02 -532.69 .8617 . = 618.18 ftDifference between Section 5 Cirrus Design Performance Data SR22 landing roll length (as Landing Distance - Flaps 100% calculated above) and WEIGHT: 3600 LB Headwind: Subtract 10% for each 13 Speed over 50 Ft Obstacle: 79 KIAS knots headwind. Flaps: 100% Tailwind: Add 10% for each 2 knots landing distance (as Power: Idle tailwind up to 10 knots. Runway: Dry, Paved, Level Runway Slope: Reference Notes Dry Grass: Add 20% to Ground Roll published in a POH) Wet Grass: Add 60% to Ground Roll PRESS DISTANCE TEMPERATURE ~C ALT FT FT 10 20 30 40 50 ISA SL Grnd Roll 1117 1158 1198 1239 1280 1321 1178 Total 2447 2505 2565 2625 2685 2747 2535 Grnd Roll 1 158 1200 1243 1285 1327 1370 1213 S = V2 / 2A = Landing 1000 Total 2506 2567 2630 2693 2757 2821 2585 Distance 2000 Grnd Roll 1201 1245 1289 1333 1377 1421 1250 Total 2568 2633 2699 2765 2832 2900 2636 . 125.062 / 2 (14.68) 3000 Grnd Roll 1246 1292 1337 1383 1428 1474 1287 Total 2635 2702 2771 284 2911 2983 2691 . 532.69 ft 4000 Grnd Roll 1293 1340 1388 1435 1482 1530 1326 Total 2705 2776 2848 2922 2996 3070 2748 According to POH Landing 5000 Grnd Roll 1342 1391 1440 1489 1539 1588 1367 distance 100% flaps with Total 2779 2854 2930 3007 3085 3163 2808 3,600 lbs at sealevel Grnd Roll 1393 1444 1495 1546 598 1649 1409 6000 Total 2857 2936 3016 3097 3179 3261 2871 . = 1,117 ft 7000 Grnd Roll 1447 1500 1553 1606 1659 1712 1453 Total 2941 3024 3108 3193 3279 3365 2937Influence of landing conditions such as runway material and surface condition, runway slope, wind conditions According to the POH the above conditions can be correct by: Dry Grass: Add 20% to ground roll Wet Grass: Add 60% to ground roll Slopped Runway: Increase table distances by 27% of the ground roll distance for each 1% of downslope. Decrease table distances by 9% of the ground roll distance for each 1% of upslope. Headwind: Subtract 10% from table distances for each 13 knots headwind. TailwindzAdd 10% for each 2 knots tailwind up to 10 knots. Grass Landings Dry Grass: Add 20% to ground roll >532.69 ft * .2 = 106 .538 ft )v 532.69 + 106.538 = 639.228 ft Wet Grass: Add 60% to ground roll 39532.69 ft * .6 = 319.614 ft r 532.69 ft + 319.614 ft = 852.304 ft Wind . Headwind: Subtract 10% from table distances for each 13 knots headwind. Headwind of 13 knots 532.69 ft - 53.27 = 480.42 ft . Tailwind: Add 10% for each 2 knots tailwind up to 10 knots. . Tailwind 10 knots 532.69 + 53.27 = 584.96 his Photo by Unknown Author is licensed under CC BY-NCSlopes Slopped Runway: Increase table distances by 27% of the ground roll distance for each 1% of downslope. Runway with 1% downslope . 532.69 ft * .27 = 143.83 532.69 ft + 143.83 = 676.52 Decrease table distances by 9% of the ground roll distance for each 1% of upslope. Runway with 1% upslope 532.69 * .91 = 484.75 This Photo by Unknown Author is licensed under CC BYInfluence of weight on landings Weight has a proportional relation to landing speeds increases in weight require and increase in landing speeds to achieve greater lift Comparing the Distance previously calculated to a lighter weight the distance required to land will decrease 3,600 lbs. requires 532.69 f't. distances 3,000 lbs. requires 436.68 ft. \fFlight Load Factor Limits 603 . Flaps Up (0%), 3600 lbs. +3.8 g, - Centrifugal Force 1.73 G's 1.9 8 . Flaps 50%, 3600 lbs. +1.9 g, 0 g - . Flaps 100%, 3600 lbs. +1.9 g, 0 g Gravity 1G Load Factor 2 G's his Photo by Unknown Author is licensed under CC BY-SStall speeds at 1, 2, 3, and maximum positive G (at 3,600 lbs 295GW . VS= 295(G) (3600) 1,062,000(G) = VS = = (CLMAX) S V(1.3408) (144.9) (194.28) . 1G 1,062,000(1) = 73.93 knots (194.28) 1,062,000(2) . 2G = = 104.56 knots (194.28) . 3G = 1,062,000(3) = 128.06 knots (194.28) 1,062,000 (3.8) . 3.8G = 144.13 knots (194.28)295L . VS = V(CLMAX) GS Calculate Stall Speed Vs= Velocity at Stall, . L=Lift = Weight = W, 3,000 . CLMAX =Coefficient of lift Max, 1.3408 . o = Density Ratio @0 ft, 1 . S = wing area, 144.9 ft2 295(3000) . VS = (1.3408) (1) (144.9) I = 67.49 knots This Photo by Unknown Author is licensed under CC BYUltimate load factor (ULF) Ultimate load factor or ULF is a 50% safety margin on the load factor This can be easily found by multiplying the maximum load factor (3.8 g) by 1.5 Resulting in 5.7 G's for the Cirrus SR22 If you multiply 5.7 by the aircraft's total weight (3,600 lbs.) positive ultimate load can be found = 20,520 lbs. Maneuvering Speed Maneuvering speed or design maneuver speed is the speed at which an aircraft can safely manipulate any one flight control once without the risk or damage to the airframe Easily found by multiplying stall speed (73.93 kts) by the square of the max load factor (3.86'5) = 144.12 The bank angle associated with the positive limiting load factor 6 5 Load factor (G units) . Using the chart on bank 4 angle you can determine that at 3.8 G there would be 3 an approximate bank angle 2 of 75% 0 10 20 30 40 50 60 70 80 90 Bank angle Figure 11.3. Load factors at various bank angles. Source: U.S. Department of Transportation Federal Aviation Administration (2008).Wecan't Rate and Radius Formula Lines . Turn rate and radius at Maneuvering Speed 144.12 kts in a 60 bank angle . Rate of turn (w) formula = ROT dea = =/Im x2+2x) 1091 Tan 8 V (ktas) xth . (2 ) 1091 Tan 60 144.12 (ktas) = 13.11 deg / sec 5 2xhth . Radius (r) formula = r = g V2 (ktas) 11.26 Tan 0 9(xth )- 9(x) hoo . (2) 144.122 (ktas) 1,065 ft S 11.26 Tan 60 = lim h/2xthThe turn performance diagram m. ; 0 Turn rate and radius at Maneuvering M 3 3 Speed 144.12 kts in a 60 bank angle _ _ f: Turn Radius (r) = approx. 1,100 ft Turn Rate at = approx. = 14 deg/sec ' L866] mm 1\" see 30 5:?" , see Banker!!! rm 'umm \fCenter of Gravity (CG) The Center of Gravity (CG) is the balance point of any aircraft REF DATUM r"'r_':' i F501) FS 10cm F51425 : i ." The datum is an imaginary line determined by the manufacture for a reference point On the Cirrus it is 100 inches forward of the firewall i la ,1; The distance from the datum to the object is called the arm 5 H22_FM06_533 The moment is calculated when the arm is multiplied by the weight and is measured in pound-inches and can be positive or negative Weight (W) * Arm (A) = Moment (M) Center of 3600 Gravity (CG) 21.1% MAC 31.5% MAC FS 143.2 FS 148.2 3600 lb 3600 lb 3400 Range Max Zero Fuel 3200 3000 CG is found by totaling all the Weight - Pounds moments of the aircraft then dividing 2800 it by the aircraft total weight. A balance point or total CG is 2600 12.5% MAC FS 139.1 2700 lb determined and can then be shifted into limits if necessary. 2400 9.8% MAC 31.5% MAC Forward limit: 2200 FS 137.8 FS 148.2 2100 lb 2100 lb 137.8 (9.8 % MAC) at 2,100 lb. to 139.1 (12.5% MAC) to 143.2 (21.1 2000 TTTTTT TITT TTTTTTTI % MAC) at 3,600 lb. 136 138 140 142 144 146 148 150 C.G. - Inches Aft of Datum SR22_EM02_5317 Aft Limit: FORWARD LIMIT - The forward limit is FS 137.8 (9.8% MAC) at 2100 lb, with straight line taper to FS 139.1 (12.5% MAC) at 2700 lb, to FS 143.2 (21.1% MAC) at 3600 lb. 148.2 (31.5 % MAC) at all weights AFT LIMIT - The aft limit is FS 148.2 (31.5% MAC) at all weights from 2100 lb to 3600 lb. from 2,100 - 3,600 lb.Center of Gravity (CG) 600 Fuel Aft Pass 500 Loading Chart Fwd Pass . CG =. Total Moment 400 Total Weight . Depending on how 300 Weight - Pounds passengers, luggage, and fuel is loaded the total CG will be 200 affected. After calculating the Baggage total CG depending on the 100 situation things may need to be shifted to bring balance 0 0.0 20.0 40.0 60.0 80.0 100.0 point within range Moment/1000 SR22_FM06_5338 . The next few slides will explain in more depthExample #1 Basic Empty Weight:2,357 lbs. Fuel: 200 lbs. PAX: Two 200 lbs. Adults Baggage: One 20 lb. bag. De-lce 20 lbs. In the first example the passenger wanted to keep his 20 lb. in the front seat to have access to snack and camera Unfortunately, this put the CG too far forward The Total CG is 140.19 This is 0.27 inches too far forward Weight 8. Balance. Normal Category (lbs. in): [WWW Arm MomI Empty Weight | 2357| 3400 |1043 133.00 3252 Front Seats ' 420 I 14350 602 Rear Seats |_ 0 | | 130.00 (Figs; = 47 gal) IWZ?\" (01.0 :28 lg 1543\" 30' Baggage | 130 [W 203.00 D-e Ice 20 | 27 | 7 181.00 3E _|l 2097 _._l 3400 __l 403 .__- 4201 3488 r 3288 , 3888 - 2888 > 2588 , Height, (lbs) 2488 - aaaaa I IVVV UVCO IIIL CIIUITYE VVILLI airspeed if a constant oV2 AOA and altitude are Lift (L) = C q s= C held? 295 Using the lift formula, if there was an increase in velocity there should be an increase in lift. VL= .1 (.8617) (100)2 295 (144.9) = 423.26lbs VL = .1 (.8617) (130)2 295 LHSV (144.9) = 715.3lbs As Velocity (V) increases so does the lift. This Photo by Unknown Author is licensed under CC BY-NCWeight & Balance, Normal Category (lbs, in): Item Weight (Ibs) Max A Arm Moment Empty Weight 2357 3400 1043 138.00 325266 Example #2 Front Seats 400 143.50 57400 Rear Seats 0 180.00 Fuel 200 486 286 154.90 30980 (tabs = 47 gal) 33.3 gal) (81.0 gal Baggage 20 130 110 208.00 4160 De-Ice 20 27 7 181.00 3620 . Basic Empty Weight:2,357 lbs. Calculate Reset Total 2997 3400 421426 . Fuel: 200 lbs. 403 140.62 . PAX: Two 200 lbs. Adults . Baggage: One 20 lb. bag. Weight and CG are within limits for the Normal category. . De-Ice 20 lbs. CG Envelope . In the second example the pilot told his 3268 passenger to keep his 20 lb. in cargo hold behind 3808 the rear seat 2808 Height . This shifted the CG 0.43 inches rearward putting 2688 the Balance Point into limits 2480 ...... . The Total CG is now 140.62 2200 .. i... 136 138 140 142 144 146 148 150 . This is within limit CG (inches)Weight & Balance, Normal Category (Ibs, in): Item Weight (lbs) Max A Arm Moment Empty Weight 2357 3400 1043 138.00 325266 Example #3 Front Seats 200 143.50 28700 Rear Seats 220 180.00 39600 Fuel 200 486 (81.0 gal) 286 154.90 30980 (tabs = 47 gal) (33.3 gal) Baggage 0 130 130 208.00 0 De-Ice 20 27 181.00 3620 . Basic Empty Weight:2,357 lbs. Calculate Reset . Fuel: 200 lbs. Total 2997 3400 403 142.86 428166 . PAX: Two 200 lbs. Adults CG Envelope 3400 . .. ..1" . Baggage: One 20 lb. bag. 3200 - . De-Ice 20 lbs. 3080 . Another option would be to have the passenger and his bag move to the back seat 2800- Height (1bs) . This shifted the CG an additional 2.24 inches 2600 keeping the Balance Point within limits 2400 . The Total CG is now 142.86 2200 - 2808 L - -. . . . . J. . 136 138 140 142 144 146 148 150 CG (inches)References Air Foil Tools. (n.d.-A). RONCZ (marske7-il) Xfoil. http://airfoiltools.com/airfoil/details?airfoil=marske7-il Air Foil Tools. (n.d.-B). Roncz/Marske-7. http://airfoiltools.com/polar/details?polar=xf-marske7-il-1000000 Air Foil Tools. (n.d.-C). HT05 (ht05-il). http://airfoiltools.com/airfoil/details?airfoil=ht05-il#polars Flying Mag. (2001). 2001 Cirrus SR22. https://www.flyingmag.com/pilot-reports-pistons-cirrus-sr22/ In Flight Pilot Training. (2016). Pilot's Operating Handbook and FAA Approved Airplane Flight Manual. https://inflightpilottraining.com/wp-content/uploads/2018/12/SR22-POH.pdf Irish, W. (n.d.). Weight & Balance. CFI Wes. http://tools.wvfc.org/WeightAndBalance.php?org=&mk=Cirrus&md=SR22&t=N809SR&ppg=&env=21 00%2C137.8%2C148.1%2C2700%2C139.1%2C148.1%2C3400%2C142.3%2C148.1&uenv=&aenv=&bas e=2357%2C138.0%2C3400%2C3409%2C3400&wma=Front+Seats%2C340%2C0%2C143.5%2CRear+Se ats%20170%2C0%2C180.0%2CFuel%3A47%2C300%2C486%2C154.9%2CBaggage%2050%2C130%2C208.0%2CDe- Ice%2C20%2C27.2%2C181.0&w1=200&w2=220&w3=200&w4=0&w5=20&cmd=Calculate Lednicer, D. (2010). The incomplete guide to airfoil usage. UIUC Applied Aerodynamics Group. https://m-selig.ae.illinois.edu/ads/aircraft.html Manuals Lib. (n.d.). Landing Distance Table - Flaps 100; Landing Distance Table - Flaps 50; Landing Distance Table - Flaps 0 -Cirrus SR22 Pilot Operating Handbook. https://www.manualslib.com/manual/1916792/Cirrus-Sr22.html?page=195#manualnow ques IIIL CianIye WILLI altitude if a constant AOA oV2 and airspeed are held? Lift (L) = C q s= C Since Density Ratio decreases as altitude 295 increases Using the lift formula, if there was a decrease in density there should be a Altitude (ft) Density ratio, Pressure ratio, Temperature (F) Temperature ratio, Speed of Kinematic viscosity, sound (kts.) v (ft2/s) decrease in lift. 0 1.0000 1.0000 1.0000 59.00 1.0000 661.7 0.000 158 1 000 0.9711 0.9854 0.9644 55.43 0.9931 659.5 0.000 161 0.9710 0.9298 0.000 165 VL = .1 (.8617) (100)2 2 000 0.9428 51.87 0.9862 657.2 (144.9) 3 000 0.9151 0.9566 0.8962 48.30 0.9794 654.9 0.000 169 295 4 000 0.8881 0.9424 0.8637 44.74 0.9725 652.6 0.000 174 5 000 0.8617 0.9283 0.8320 41.17 0.9656 650.3 0.000 178 = 423.26lbs 6 000 0.8359 0.9143 0.8014 37.60 0.9587 647.9 0.000 182 7 000 0.8106 0.9004 0.7716 34.04 0.9519 645.6 0.000 187 8 000 0.7860 0.8866 0.7428 30.47 0.9450 643.3 0.000 192 VL = .1 (.7385)(100)2 9 000 0.7620 0.8729 0.7148 26.90 0.9381 640.9 0.000 197 295 (144.9) = 10 000 0.7385 0.8593 0.6877 23.34 0.9312 638.6 0.000 202 15 000 0.6292 0.7932 0.5643 5.51 0.8969 626.7 0.000 229 362.74lbs 20 000 0.5328 0.7299 0.4595 12.32 0.8625 614.6 0.000 262 25 000 0.4481 0.6694 0.3711 -30.15 0.8281 602.2 0.000 302 As pressure (o) decreases 30 000 0.3741 0.6117 0.2970 47.98 0.7937 589.5 0.000 349 so does the lift. 35 000 0.3099 0.5567 0.2353 -65.82 0.7594 576.6 0.000 405 36 0894 0.2971 0.5450 0.2234 -69.70 0.7519 573.8 0.000 419 40 000 0.2462 0.4962 0.1851 69.70 0.7519 573.8 0.000 506 45 000 0.1936 0.1455 -69.70 0.7519 573.8 0.000 643 50 000 0.1522 0.3002 0.1145 -69.70 0.7519 573.8 0.000 818How do the required CL and ADA for your specific aircraft (at a specific weight) change with changes in airspeed? Based on the previous slide we determined that if there is an increase in airspeed there will be an increase in lift. ADA and CL are positively correlated. In order to maintain level flight, the required CLand therefore AOA will need to be reduced ifthe airspeed increases. Lift (L) = . g .. .2 '1 L = _1 Manny) =423.25ibs i.3617)i_110.12 ' 423.23 lbs 2 x (144.9) / To not increase lift the ADA (x value) must decrease. J 423.23 lbs : x 5121.39 '/ (423.23} _ 5121.39 _ x \\/ Reducing the ADA to .083 while increasing airspeed to 110 knots, the Aircraft will maintain level flight. x: .083 What happens if the required CL is larger than CLIVIAX of your airfoil, and what speed regime is usually associated with . L: that condition? Cl v Alpha \"Since the CL MAX is the maximum AOA before stalling a higher CL is not possible. "40 '55" "5999 \"If a greater lift was required \"0 -3.75 ~3474 one would need to increase "0\" 0 .0425 airspeed to maintain the \"C" climb or prevent the aircraft 0-60 2 376 from stalling. 0.40 4D .6284 . . m \"Alternatively, the pilot could ' 6 3652 descend to increase density 0.00 . an 1.0673 altitude and therefore \"0 increase lift. 70.40 10 1.1625 @760 -10.0 -5.0 CLD 50 10.0 15.0 ELM 120 1.2317 140 1.3068 150 1.3361