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\fNon-symmetrical vs Nonsymmetrical Symmetrical . In a cambered airfoil ( nonsymmetrical), the mean camber line and Symmetrical chord line are different. If the mean camber line coincides with the chord line, the airfoil is said to be symmetrical. . The NACA 2412 is a non- symmetrical airfoilAOA / CL . The NACA2412 airfoil does not have an exact Zero-Lift AOA, at -0.0003 CL, alpha is -2.205. XFOIL Version 6.96 N71205 Calculated polar for : NACA 2412 1 1 Reynolds number fixed Mach number fixed xtrf = 1.000 (top) 1.000 (bottom) Mach = 0. 000 Re = 1.000 e 6 Ncrit = 9.000 Details Airfoil: NACA 2412 (naca2412-il) alpha CL CD CDp CM Top_XtraBot_Xtr Reynolds number: 1,000,000 Max CI/Cd: 101.38 at a=4.5 Description: Mach=0 Ncrit=9 -3. 750 -0. 1645 0. 00752 0 . 00205 -0. 0553 0. 8758 0. 1310 Source: Xfoil prediction - 3. 500 0. 1372 0. 00737 0 . 00192 -0. 0551 0. 8645 0 . 1482 Download polar: xf-naca2412-il-1000000.txt - 3. 250 0. 1100 0. 00719 0. 00180 -0. 0549 0. 8525 0. 1711 Download as CSV file: xf-naca2412-il- - 3. 000 -0. 0825 0 . 00704 0 . 00170 -0. 0547 0. 8404 0. 1957 1000000.CSV - 2. 750 -0 . 0552 0. 00689 0 . 00160 -0. 0546 0. 8278 0. 2215 - 2. 500 -0. 0277 0. 00678 0 . 00152 -0. 0544 0. 8151 0. 2471 - 2. 250 -0. 0003 0. 00666 0 . 00146 -0. 0542 0. 8016 0. 2776 000 7- 0. 0272 0. 00653 0 . 00140 -0. 0541 0. 7879 0. 3097 -1. 750 0. 0546 0 . 00640 0 . 00134 -0. 0540 0. 7740 0. 3442 - 1. 500 0 . 0819 0 . 00628 0. 00129 -0. 0538 0. 7597 0. 3795 -1. 250 0. 1092 0. 00616 0. 00125 -0. 0536 0. 7451 0. 4198 - 1. 000 0. 1362 0. 00602 0. 00122 -0. 0534 0. 7296 0. 4712 -0. 750 0. 1632 0. 00589 0. 00121 -0. 0532 0. 7137 0. 5267 -0 . 500 0. 1903 0. 00580 0. 00120 -0.0530 0. 6972 0. 5727 -0. 250 0. 2173 0. 00573 0. 00121 -0.0527 0. 6799 0. 6158Cl v Cd Cl v Alpha 2.00 2.00 Polar Graphs 1.50 1.50 1.00 1.00 0.50 0.50 Cm v Alpha 0.00 0.00 0.00 -0.50 -0.50 -0.01 -1.00 -1.00 -0.02 -1.50 -1.50 0.03 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 -20.0 -15.0 -10.0 -5.0 0.0 5.0 10.0 15.0 20.0 Cl/Cd v Alpha Cd v Alpha -0.04 120 0. 10 100 0.09 -0.05 80 0.08 -0.06 0.07 60 0.06 -0.07 40 -20.0 -15.0 -10.0 -5.0 0.0 5.0 10.0 15.0 20.0 0.05 20 0.04 0 0.03 -20 0.02 -40 0.01 -60 0.00 -20.0 -15.0 -10.0 -5.0 0.0 5.0 0.01 10.0 15.0 20.0 -20.0 -15.0 -10.0 -5.0 0.0 5.0 10.0 15.0 20.0W=25501bs (MOTW) Stall Speed CLmax=1. 5820 Stall Speed at maximum take off weight 25501bs. Density Ratio o =1 Vs = V295 (W) / (CLmax) (o) (S) S=wing area 174 ft^2 Vs=(295) (2550) / (1.5820) (1) (174) Vs = 52.3 KtsHow does lift change airspeed if a constant ADA and Altitude are held? Bernoulli's principle states that an increase in velocity will cause a decrease in pressure. Therefore, we can conclude, that lift is generated by an airfoil based off of how fast air moves over as well as under the surface of said airfoil. Bernoulli's theorem states that the increased speed atop the wing (airfoil) is associated with a region of lower pressure. which is lift. when a Constant ADA and altitude are maintained but the airspeed changes, the change results in a pressure change which will result in a change in Lift. Formula for Lift(L): (6)0\") L = (CL) 295 (S) . L = (CL)- (0) (V2 ) - (S) 295 . At an altitude of 10, 000 feet traveling at 120 knots. Table 2.1 of the text o = .7385. Example: 120 vs 100 CLmax=1.5820, AOA 16.5. Our wing area is 174 ft^2. . L = (1.5820) (.7385) (1202 (174) 295 . L = (CL) ( 0 ) (V 2 ) . L = 9923.1 lbs (S ) 295 . At an altitude of 10,000 feet traveling at 100 knots. Table 2.1 of the text o = .7385. CLmax=1.5820, AOA 16.5. Our wing area is 174 ft^2. . L = (1.5820) (.7385) (1002 ) (174 ) 295 . L = 6,891.0 lbs By changing the airspeed there is a significant difference in lift- The Density Ratio (0) will change as the aircraft descends or climbs which changes altitude. However, the ADA and V will stay the same. (0)(V2) 295 Using the same variables from our previous slide at 128 knots and Table 2.1. L 2 (CL) (S) . L = (15820)EZEE1352(174) 295 ' L = 9923.]. lbs Now the aircraft descends to 5009ft MSL. Table 2.1: a=.8617. with the same ADA and V the result is more lift because the air is more dense at lower altitudes. More lift is generated. (.8617)(12 - L=(1.5820) 295 @074) ' L = 11578.5 lbs As speed decreases CL increases How do the required CL and AOA for the Cessna 172 at 2550 lbs change with change in airspeed? . CI = (295) (W) (0 ) (V2 ) (5 ) . CL = (295) (W) (0 ) (V2) (5 ) . Example 120knots at sea . Example 100knots at sea level standard day. level standard day. . CL = (295) (2550) (295) (2550) (1) (1202 ) (174 ) . CL= (1)(1002) (174) . CL =0.3002 . CL = 0.4323l l Wha_t happens if the required CL is larger than the CLmax of the airfoil? What Speed regime is usually associated with that condition? - An accelerated stall will be caused if the CL is larger than the Clmax. \fDrag: Drag is the opposing force to thrust. It is caused by aerodynamic resistance as an object moves through the air. Drag is generated by the interaction of a solid body with a fluid (liquid or gas), and for drag to be generated, the solid body must be in contact with the liquid or gas. when it comes to aerodynamic drag on an aircraft is an umbrella term. It is the sum of several different types of drag that have different sources. - Induced Drag: Di, is the least understood type of drag, but it is the most important, especially in the critical low-speed region of flight. It is called the drag due to lift because it occurs only when lift is developed. Induced drag occurs because the distribution of lift is not uniform across the wing as it varies from the wing root to the wingtip. Induced drag is influenced by the lift coefficient and aspect ratio. It increases directly as the square of CL and inversely with aspect ratio. - Parasite Drag: DP, is easily understood, but difficult to measure as it is the drag that is not associated with the production of lift. As speed increases, the induced drag decreases, but parasite drag will increase because the \"solid body\" is moving through the \"fluid\" with greater force and thus experiences the force of drag in a greater capacity. Parasite Drag + Induced Drag = Total Drag iii 810 820 150 2180 2650 5940 810 330 1385 Construction of Drag 0. 02547 740 970 870 760 830 -1030 40 Assumptions : 150 2810 380 2560 Chart: . Cessna 172 (NACA 2412 1250 880 710250 . Weight (W) : 2550 1bs 155 425 . Wing Area (S) : 174 ft2 270 . Aspect Ratio (AR) : 7.45 3175 1125 250 1185 2500 . CLmax: 1.582 . CDP : 0. 02547 3100 . Efficiency Factor: 0.82 5 6 Atmospheric conditions: 200 . Sea level . Temperature: Standard 15015 1200 1000'01 Secant Lines Tensew Terminology (as used in Drag Chart) hide +(X)= lim (xthkx T h = lim: x2+2xhth?-x2 q(psf)= Dynamic Pressure CL= Coefficient Lift xth S CD = Coefficient of Drag 2xWith ? Cop =Coefficient of Parasite Drag = lim 9(xth)- 9(x) hoo CDi =Coefficient of Induced Drag Dp = Parasite Drag Di = Induced Drag DT = Total Drag\fRequired formulas (continued) Calculate CDP: Referencing CDP from airfoiltools.com CDP for the airfoil: 0. 00547. add 0.02 to the airfoil CDP which will account for the parasite drag. Total CDP = 0. 02547 . CDI = [1 *e*AR] *CL^2 = [1 / (n * 0.82 * 7.45) ] * 1.582^2 = 0.1304 . CD = CDP + CDI = 0. 02547 + 0. 1304 = 0.15587 . CL / CD = 1.582 / 0.15587 = 10.15 . DP = CDP * q * S = 0. 02547 * 9.27 * 174 = 41.08 . DI = CDI * q * S * DI = 0. 1304 * 9.27 * 174 = 210.33 . DT = DI + DP = 41. 087 + 210.33 = 251 .417weight 2550 wing area (S) 174 Raw Data Aspect Ratio 7.45 0.82 CDP 0.02547 Clmax 1.582 Constant 295 V q (psf) CL CDP CDI Co CL / Co DP (lb ) Di(lb) DT ( lb ) 52.3 9.27 1.581 0.02547 0.1302 0.1556 10.16 41.09 210.01 251.10 72 17.57 0.834 0.02547 0.0362 0.0617 13.51 77.88 110.81 188.69 92 28.69 0.511 0.02547 0.0136 0.0391 13.08 127.15 67.87 195.02 112 42.52 0.345 0.02547 0.0062 0.0317 10.89 188.45 45.79 234.24 132 59.06 0.248 0.02547 0.0032 0.0287 8.65 261.76 32.97 294.73 152 78.32 0.187 0.02547 0.0018 0.0273 6.86 347.09 24.86 371.95 172 100.28 0.146 0.02547 0.0011 0.0266 5.50 444.44 19.42 463.86 192 124.96 0.117 0.02547 0.0007 0.0262 4.48 553.81 15.58 569.39 212 152.35 0.096 0.02547 0.0005 0.0260 3.71 675.19 12.78 687.97 232 182.45 0.080 0.02547 0.0003 0.0258 3.11 808.60 10.67 819.27 252 215.27 0.068 0.02547 0.0002 0.0257 2.65 954.02 9.05 963.07 272 250.79 0.058 0.02547 0.0002 0.0256 2.28 1111.46 7.76 1119.22 292 289.03 0.051 0.02547 0.0001 0.0256 1.98 1280.92 6.74 1287.66Drag Curve Chart 1400.00 1200.00 Analyzing the data 1000.00 800.00 What are the minimum drag 600.00 parameters for the Cessna 172: . Minimum drag (DMin) : 188. 69 400.00 . Speed VD (min) at Dmin : 72 KTAS 200.00 Relationship between Do and D, at Dmin B 0.00 52.3 72 92 1 12 132 152 172 192 212 232 252 272 292 Although the parasite Drag, DP (lb) DI (lb) DT (Ib) (Dp) is higher than the induced drag(Di ), at the minimum drag value D(min) , DP and DI are closest together at D(min)what are the maximum lift to drag ratio CL/CD parameters of the Cessna 172?: Analyzing the data I - (Ci/CD)max Value: 13.51 (cont'nued) - Speed at which (CL/CD)max occurs: 72 KTAS Drag Curve Chart Comparing the min and max parameters: It appears that the most efficiency for the airfoil is achieved at 72 KTAS Glide performance: To predict glide performance we can divide the CL by the CD Glide Performance : Ci/CD (CL) 0.834 / 0.6617(CD) = 13.51 Pilots usually refer to glide performance with a glide ratio: Glide Ratio = Horizontal Distance/Change in altitude or how far did the glider travel forward for every foot it dropped in altitude 730.00 In an emergency scenario where the engine has lost power, pitching for best glide (65 knots) would give the best glide distance. - Glide Ratio : 1.5 nautical miles per 1,666 feet. - At 18,898 feet pitching for best glide of aircraft could glide about 15nm. ' 1.5*16=15nm \fFocus Focus concepts of unaccelerated performance: - Climb performance - Cruise performance - Glide performance why are these important? Pilots should always understand the characteristics of their aircraft so they can harness its performance to its most optimal; when conditions allow for positive performance. Alternately, understanding performance when conditions negatively impact performance. Required formulas - To Calculate power required we can utilize the formula ' PP =D*Vk/325 ' PP=251.16*52.3/325 - Pr=40.41 Constructing data chart/Raw Data Pr 2:253:25? e cop Clm ax q (psf) o 0...: cm 0., c. / on 0505) D.(|b) 0.00) Pr (HP) Constant 9.27 1.581 0.02547 0.1302 0.1555 . 41.09 210.01 251.10 40.41 17. 57 0. 834 0. 02547 0. 0352 0.0517 . 77.88 110.81 188.59 41.80 28. 59 0 511 0 02547 0 0135 0.0391 . 127.15 57.87 195.02 55.21 59. 05 0 248 0. 02547 0 0032 0.0287 . 251.75 32.97 294.73 119.70 0 02547 _ . _ _ _ 173.951 0.02547 . . . . . . 245.49 0.02547 . . . . . _ 335.38 . 0.02547 . . . . . . 448.77 - 0.080 0.02547 0.0258 . 808.50 584.83 215 27 0.058 0.02547 0.0002 0.0257 . 954.02 9.05 953.07 745.75 250 79 0.058 0.02547 0.0002 0.0255 2.28 1111.45 7.75 1119.22 935.70 289. 03 0.051 0.02547 0.0001 0.0255 1.98 1280.92 5.74 1287.66 1155.91 Analysis - When total drag increases as airspeed is increasing, more power is required to maintain the airspeed. However, more power is required when airspeed falls below stall speed, because induced drag increases rapidly. ~Power requirement is the lowest at 7292 knots, hence takeotf is achievable at those speeds since the 172 has a max takeoff power of 180 HP. Air Speed vs Power Required 1400.00 1200.00 1400.00 Drag Vs Power Required 1000.00 1200.00 800.00 1000.00 600.00 800.00 400.00 600.00 400.00 200.00 212.00 232.00 252.00 272.00 112.00 132.00 152.00 172.00 192.00 200.00 0.00 52.30 72.00 92.00 -Pr (HP) 0.00 40.41 41.80 55.21 80.72 119.70 173.96 245.49 336.38 448.77 584.83 746.75 936.70 DT (1b) 1 156.91 Pr (HP) Analysis (Continued)Maximum range ahspeed ' The maximum range condition is obtained at maximum lift/ drag ratio (L/DMAX), and it is important to note that for a given aircraft configuration, the L/DMAX occurs at a particular ADA and lift coefficient and is unaffected by weight or altitude. L/D Max is at 13.72, 80 knots for the Cessna 172 q (psf) CL COP CDI CL / Co DP (lb) DI (1b) D. (lb ) Pr (HP) 52.30 9.27 1.581 0.02547 0.1302 0.1556 10.16 41.09 210.01 251.10 40.41 60.00 12.20 1.201 0.02547 0.0751 0.1006 11.94 54.08 159.56 213.65 39.44 65.00 14.32 1.023 0.02547 0.0546 0.0800 12.79 63.47 135.96 199.43 39.89 70.00 16.61 0.882 0.02547 0.0406 0.0660 13.3 73.61 117.23 190.84 41.10 75.00 19.07 0.769 0.02547 0.0308 0.0562 13.66 84.50 102.12 186.62 43.07 80.00 21.69 0.676 0.02547 0.0238 0.0492 13.72 96.15 39.75 185.90 45.76 85.00 24.49 0.598 0.02547 0.0187 0.0441 13.56 108.54 79.51 188.05 49.18 90.00 27.46 0.534 0.02547 0.0148 0.0403 13.24 121.69 70.92 192.60 53.34 95.00 30.59 0.479 0.02547 0.0120 0.0374 12.80 135.58 63.65 199.23 58.24 100.00 33.90 0.432 0.02547 0.0097 0.0352 12.28 150.23 57.44 207.67 63.90 120.00 48.81 .300 0.02547 0.0047 0.0302 9.95 216.33 39.89 256.22 94.60 125.00 52.97 0.277 0.02547 0.0040 0.0295 9.39 234.73 36.76 271.50 104.42 130.00 57.29 0.256 0.02547 0.0034 0.0289 8.86 253.89 33.99 287.88 115.15- Maximum endurance is defined on the time an Maximum endurance aircraft can remain in flight airspeed - Maximum endurance airspeed is the stage where an aircraft utilizes minimum fuel power. - As shown on the previous slide, the maximum endurance happens at (L/D)max which is shown to be at 86 kts Best climb conditions Best rate of climb (ROC) & associated airspeed - Best angle of climb (AOC) & associated airspeed Best Rate of Climb . Best rate of climb is defined as the ability to gain the greatest ROC=33000 [ (180-13.72) ]/2550 amount of altitude in the shortest time, by maximizing ROC= 2, 152 fpm climb airspeed. Formula : ROC=33000 (Pa-Pr) /W Known : . Pa 180 1bs . Pr 13.72 (80 kts L/Dmax) . Aircraft Weight W 25501bssing= 325 (180-41.8) /72*2550 sing=3.1 Degrees Best Angle of Climb . Best Angle of Climb is defined as the pitch angle at which the aircraft can gain the greatest amount altitude in the shortest distance. Formula . AOC= siny= 325 (Pa-Pr) /VKW Known . VK 72 . Pa 180 . Pr 41.8 . W 2550Maximum forward According the drag table, at about 152 kts, total drag is 371.95 lbs, thus will require about 180 horsepower, for which we know is the maximum allowable power produced by the 172 engine. Therefore, we will not be able to achieve any higher speed in level flight. Max Forward Speed is where Pa=Pr Best Glide Airspeed is the airspeed at which the aircraft can cover the most distance horizontally with lowest decent speed. To predict glide performance we can divide the CL by the CD Glide Performance = CL/CD (Ci) 9.676 / 6.6492(CD) = 13.72 Looking at the drag chart, between our stall speed of 52 and first airspeed of 72 which shows the possibility of being airborne with lowest drag. By adding both these values together and dividing by 2 we get an airspeed of 62.15 kts. - The Cessna 172 has a published Vg of 65 knots - Glide Ratio = 1.5 nautical miles per 1,966 feet. - At 16,966 feet pitching for best glide of 65 the aircraft could glide about 15nm. 1.5*16=15nm \fFactors that effect Take-off /Landing Aircraft Weight Thrust Available C-GWKB Ambient Temperature GWKS Atmospheric density or pressure Wind speed and direction Runway inclination or declination Runway surface conditions6000 Takeoff speed Weight Associated conditions pounds Liftoff 50 ft Power Full throttle 2600 rpm kts MPH kts MPH 000 Mixture Learn to appropriate 2950 66 76 72 83 Reference line fuel pressure Reference line Reference line Guide lines Intermedia 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 Tailwind 2200 58 67 63 73 Cowl Open flaps 3000 Headwind 10/000_Pressure altitude - feet ISA 2000 8000 6000 1000 4 000 2000 S.L. 0 C -40 -30-20-10 0 10 20 30 40 50 2800 2600 2400 2200 0 10 20 30 0 50 Outside air temperature Weight Wind component Obstacle (lb) (knots) height (ft) F -40 -20 0 20 40 60 80 100 120 Take-off (utilizing figure 8.1) Figure 8.1. Takeoff distance graph.Using Figure 8.1 we can determine take off speeds based off of changing factors such as temperature, weight, Takeoff roll based on wind component and air density. Scenario 1: Class alpha aircraft trying to takeoff at environment: weight of 2, 600 1bs., Pressure Altitude 2000, and temperature of 30 Celsius. 5 knot headwind and 50 ft obstacle: 6000 Tracing the table to the set weight and then looking Takeoff speed Weight Associated conditions to the right under the "Lift-Off" column, it will pounds Liftoff 50 ft Power Full throttle 2600 rpm kts MPH kts MPH 000 require no less than 63 kts to be able to takeoff. Mixture Learn to appropriate 2950 66 76 72 83 Reference line Reference line fuel pressure Reference line Guide lines Intermedi Following the Red line tracing the scenario it will 2800 64 74 70 81 Flaps Up 2600 require 1200 ft takeoff distance. 63 72 68 78 Landing Retrack after positive 4000 Obstacle heights 2400 61 70 66 76 gear climb established Tailwind 2200 58 67 63 73 Cowl Open Scenario 2: Class alpha aircraft trying to takeoff at flaps 3000 weight of 2, 600 1bs., Pressure Altitude 2000, Headwind temperature of 30 Celsius. 5 knot tailwind and 50 ft obstacle: 19000 Pressure altitude - feet ISA 2000 Following the orange line tracing the same scenario 8000 with a 5 kts tailwind, it will increase to 1800 ft. 6000 1000 4000 2000 S.L Notice that as headwind increases and temperature decreases, takeoff roll decreases, since air density C -40 -30 -20 -10 0 10 20 30 40 50 2800 2600 2400 2200 0 10 20 30 0 50 increases, increasing the lift component, allowing Outside air temperature Weight Wind component Obstacle an early efficient takeoff. The opposite occurs when (lb) (knots) height (ft) F -40 -20 0 20 40 60 80 100 120 these factors are reversed. Figure 8.1. Takeoff distance graph. Hence environment (in this case wind) effects take off distance.Landing Factors l0! Ii Weight wind direction Ambient and speed Temperature Ax \\/ Density Landing Altitude Configuration Landing approach Speed The best landings happen when energy management is achieved successfully with the landing environment Factors considered. Scenario: - wind calm - Temperature/density is at sea level 15/29.92 - Weight 2,550 - Given, Vs at 2556 lbs. = 52.3 kts. Approach speed is 1.2 times the stall speed ' Vref = 52.3 X 1.2 = 62.76 kts - Best approach speed is 63 kts Total Drag /Average Drag at Touchdown Con Co: '0.) 'Ct I Co D(Ib) D-(lb) Dv(|b) Pr (HP) 1.581 0.02547] 0.1302L 0.1556] 10.16 41.091 210.01 251.10 40.41 - Total drag at touchdown is equal to the total drag at stall speed - For the Cessna 172 s, with 2,556 lbs, total drag at touchdown is 251.10 lbs. - Drag decreases as the aircraft slows down during landing r011, parasite drag increase with increasing speed, therefore it decreases as the aircraft decelerates. - Assuming the aircraft will decelerate in a constant and linear manner; runway slope is level, no wind gust, and runway condition is uniform. The average drag during landing roll should half the total drag at touchdown, which in this case is about 139 lbs. ' 251.191bs / 2 = 125.551b5 q (psf) C / Co DP (1b Di (lb Or (lb Pr (HP 52.3 9.27 1.581 0.02547 0.1302 0.1556 10.16 41.09 210.01 251.10 40.41 72 17.57 0.834 0.02547 0.0362 0.0617 13.51 77.88 110.81 188.69 41.80 92 28.69 0.511 0.02547 0.0136 0.0391 13.08 127.15 67.87 195.02 55.21 Residual thrust Residual power based on the rated power of 180 hp Hp / max RPM . 180 /2400 = 0.075 Rated power factor * assumed landing rpm . 0. 075* 700= 52.5 HP Thrust at touchdown 325*HP /V . Thrust =(325*52.5) /72 . Thrust=236.97 -188.69 = 48. 291bs Average thrust during the landing roll (325*52.5) /52.3 = 326.24 (326. 24-251. 10) +48.29/2 = 61.72 1bsDeceleration . To calculate the aircrafts deceleration we need to take into consideration the aircrafts mass, residual thrust, drag and friction from the nose and main wheel. . To calculate the mass we will use the equation: - m=WT/32.2 - Landing Weight 2550 m=2550/32.2 = 79.19 slugs . To find the friction we use the equation - Fr=. 02* . 25*WT for front wheel The equation for Deceleration is : a=T-D-Fr-Fb/m Fr=. 02* . 25*2550= 12.75 - Fb= . 73* . 75*WT for main wheels a=0-125.55-12.75-1396.13/79.19 Fb= . 73*.75*2550= 1, 396.13 1bs a=-19.38 ft/s^2 Drag has been calculated at 125.55 (Slide 42)X Landing Roll Landing Scenario: . Runway condition: dry Length of the landing roll Multiple . Runway Slope: 0 Degrees stall speed (V) 52.3 x 1.2 = 62.76 (level) . Obstacles : N/A . Wind : Calm S=v^2/2a . Aircraft Weight 2550 Converting velocity to fps 63 x 1.69= 106.47 fps. S=(106.47) ^2/ 2(19.38) S=11, 335 . 86/38.76 = 292.46 ftThe Role of Friction in Landing Aerodynamic Braking: Holding the nose up, increase resistance against the airflow, works the same principle as parasite drag, and in turn sows down the aircraft. Wheel braking: restrict tire rotation, thus slowing down the aircraft. Note: inertia will force the nose down while using wheel braking, therefore nose up aero dynamic braking cannot be used at the same time as wheel braking. Influence of altitude on landings - Altitude effects aircraft performance and this also applies, to landing performance particularly landing distance. - Higher density altitude means a higher landing speed - Lower density altitude reduces landing speed - Therefore, the Cessna 172 will require a longer runway distance for mountainous destinations and can suffice with a smaller runway for sea level destinations depending on environmental factors. CESSNA SECTION 5 MODEL 172S PERFORMANCE SHORT FIELD LANDING DISTANCE AT 2550 POUNDS Influence of CONDITIONS: Flaps 30 Power Off Maximum Braking Paved, level, dry runway Zero Wind Speed at 50 Ft: 61 KIAS altitude on 0.C 10 20 C 30 C 40 Grnd Total Grnd Total Grnd Total Grnd Total Grnd Total Press Roll Ft To Roll Roll Ft To Roll Ft To Ft To Alt Ft Clear Ft Clear Ft Clear Ft Clear Ft Clear 50 Ft 50 Ft 50 Ft 50 Ft 50 F landings cont.. In Obst Obst Obst Obst Obst Feet S. L 545 1290 565 1320 585 1350 605 1380 625 1415 100 565 1320 585 1350 605 1385 625 1420 650 145 2000 585 1355 610 1385 630 1420 650 1455 670 1490 3000 610 1385 630 1425 655 1460 675 1495 695 1530 Notice as the pressure altitude and 4000 630 1425 655 1460 675 495 700 535 725 1570 5000 655 1460 680 1500 705 1535 725 1575 750 1615 temperature increases the ground roll and 6000 680 500 705 1540 730 1580 755 1620 780 1660 obstacle clearance needed increases as well. 7000 705 1545 730 1585 760 625 785 1665 810 1705 8000 735 1585 760 1630 790 1670 815 1715 840 1755 PA: Sea level at 20 degrees: NOTES: Ground roll: 585 ft, 50ft obst clear : 1350 ft 1. Short field technique as specified in Section 4. 2. Decrease distances 10% for each 9 knots headwind. For operation with tail winds up to 10 knots, increase distances by 10% for each 2 . PA: 4000 ft at 20 degrees: knots. Ground roll:675 ft, 50 ft obst clear : 1495 ft 3. For operation on dry, grass runway, increase distances by 45% of the "ground roll" figure. 4. If landing with flaps up, increase the approach speed by 9 KIAS and allow for 35% longer distances. PA: 4000 ft at 30 degrees: Figure 5-11. Short Field Landing Distance - Ground roll: 700 ft, 50 ft obst clear : 1535 ft Revision 4 5-23/5-24Difference between landing roll length: (as calculated above) and landing distance (as published in a POH) Influence of landing conditions : (such as runway material and surface condition, runway slope, wind conditions) Calculated landing distance 292.46 ft (dry runway condition) POH states to add 45% to ground roll for dry grass runway 292.46 * .45: 131.607 'F't 292.46 + 131.607: 424.43 'Ft \f3500 weight and Associated conditions Weight Speed at 50 ft Power Retarded to maintain (lb) Guide lines for MPH applicable for Intermediate 3000 kts. 900 fi/on final approach Obstacle heights Reference line Reference line Flaps Down 2950 70 Reference line Landing gear Down 2800 58 78 Runway Paved, level, dry surface 2600 65 75 2500 landings Approach speed IAS as tabulated 2400 63 72 allwind Braking Maximum 220 60 69 2000 . When aircraft weight is increased Pressure altitude (ft) 1500 so is the speed to maintain lift. 10 600 Therefore, the landing distance 8000- dwind 6000 4000 1000 will increase as well. 20007 SL You can see at 24001bs the red 500 line only requires 900 ft of C-40-30 -20-10 0 101 20 30 40 50 2800 2600 2400 2200 0 10 20 30 0 50 ground roll and the yellow line Outside air temperature Weight Wind component Obstacle (lb) (kts. height (ft) at 26001bs (no other element F -40-20 0 20 40 60 80 100 120 changed) requires 1100 ft of ground roll. Figure 9.1. Landing distance graph. As weight increases so does speed and landing distance.V-G Diagram Data MTOW: 2550 1bs CL max = 1.581 Aircraft Wing Positive Load Area: 174 ft2 factor +3. 8G Stall Speeds Negative Load 1G= 52 Knots Factor - 1.52G Sea level 2G= 73.6 Knots 3G= 90.1 KnotsLn [I 'vu 40 5o Load Factor ED L OD 122 HE 160 15.5 V-G Diagram- cont - VG Diagram shows the stall velocities within the load factor range. - The aircraft must fly within the speed and G ranges to the right of both curves. Anything to the left of them and the aircraft will stall. \fLoad Factor The definition of load factor in aeronautics is the ratio of lift of an aircraft to its weight. Since the load factor is the ratio of two forces, it is dimensionless. However, its units are traditionally referred to as g, because of the relation between load factor and apparent acceleration of gravity felt on board the aircraft. A load factor of one, or 1 g, represents conditions in straight and level flight, where the lift is equal to the weight. Load factors greater or less than one (or even negative) are the result of maneuvers or wind gusts. When the load factor is greater than +1 all occupants feel heavier than usual. Examples: in a 2 g maneuver all occupants feel that their weight is twice normal. When the load factor is zero, or very small, all occupants feel weightless. When the load factor is negative, all occupants feel that they are upside down. eed (D O a m g .E m m m 9 o .E - c m 9 m D. aims Load factor or \"G" OAIU w L m m N m m 1 5 Increase in stall speed with load iacmr. Em, Positive and Negative Limits 0 Positive and negative limits of a wing will both lead to a stall. This is a result of wing loading by G force created in dynamic flight. Turns specifically increase the load on a wing, maximizing usable lift. An increase in thrust is what is used to maintain level flight in a turn. However, this places additional load on the wing and can push a wing to its limit, resulting in a stall. A load factor greater than 1 will cause the stall speed to increase by a factor equal to the square root of the load factor. Example: if the load factor is 2, the stall speed will increase by, or about 40%. Stall Speeds - To calculate the stall speed at different load factors, multiple the stall speed at (52 knots) by the square root of the load factor. Cessna 172 16: 52*: 52 knots Weight: 2559 lbs Sea level 26:52*V2==736knm3 Stall Speed Calculated: 52.3 or 52 Load Factor @ 52 = 16 26: 52*\\/ = 90.1 knots 3.8 G maximum Load Factor 52*V3.8 = 101.4 knots Stall Avoidance AVOID FLYING AT REMAIN IN THE NORMAL AVOID ABRUPT MINIMUM AIRSPEEDS FLIGHT ENVELOPE MANEUVERS \fULF Ultimate load factor (ULF) is found by taking the Max load factor (3.8 G) and multiplying it by 1.5: 1.5 * 3.8 Positive Ultimate Load Weight of Ultimate Load Factor: Ultimate Load Factor(5.7 G) multiplied by Aircraft Weight (25501bs) 5.7 * 2550 = 14,535 lbs Maneuvering speed Maneuvering speed VA: The speed at maximum G-limit Stall speed (52 knots ) multiplied by VA = VSVLLF the square root of max load factor (3.8 where G) VA = maneuver speed (kts.) Vs = stall speed (kts.) VA =52(V3.8)= 101.4 Knots LLF = limit load factor . Pilots need to heed VA not only when performing acrobatic maneuvers and steeply banked turns, but should also slow the aircraft to an airspeed below VA when entering turbulent air.Load taste: (6 arms} M ' ' m 1a 20 30 40' 50' 60' ,0, an Eankangle hm mm mm m \"mun, wk and\" 90' The bank angle associated with the positive limiting load factor. - In order to Find the bank angle associated with the positive limiting load Factor (3.8 6) OF the Cessna 172, we use the following formula: ' (D : cos1(l/G) ' (D = cos1(1/3.8) ' 0 : 74-.7424-77n - (D = 75 - Therefore, at a 75 bank (turn), in order to maintain level Flight the aircraFt will be pulling 3.8 Gs. Any higher bank and the aircraft will need to start descending, or the positive limit load factor will be exceeded. Turning Rate & Radius Rate of turn is the rate of change of direction; how many degrees are turned through in a specific timeJ usually a minute. Radius of turn is the size of the arc made by the aircraft as it turns. A low speed means a higher rate of turn; a higher forward speed means a lower rate of turn. The radius of a turn is critical to the success of a maneuver, whether it be air-toair combat or simply turning base to final. Increased bank angle during a coordinated turn in level flight decreases the turn radius. when an aircraft increases its bank angle to limit the turn radius in level flight, stall speed increases. A balance must be found between the bank angle necessary to perform the maneuver and the need to maintain a safe margin above stall speed without exceeding VA. There are two methods to finding rate and radius, mathematical and constant altitude turn performance charts. Rate of turn (ROT): To calculate the Rate 0? Turn (ROT) we will use the constant 1091 and multiply it by the bank angle, then divide the product by the given airspeed in knots. 7 109104271123) _ VK ROT Example: 5 36\"bank at our maneuvering speed of 161.4 knots, results in a Rate of Turn of 5.9 per second. _ 1091(tun30) _ 101.4 TURN COORDW ROT = 6.2\" 2 MIN. NO PITCH WFOPMN'MDN Mathematical Method (radius) The turn Radius (r) is calculated using the Following equation: V1? r : 11.26 tuna In order to find the radius of the turn, we square the maneuvering speed (161.4 knots) and divide it by the tangent of the bank angle (tan3edeg) multiplied by 11.26. 7 (101.4)2 f _ 11.26 (mam) r : 1581.6 ft '///. - Using figure 11.14 the text, we can determine our rate of turn and radius. - Knowing our maneuvering speed of the aircraft which we have determined is 161.4 knots, we can plot the 100 knots mark at the bottom of the graph and follow it up until we intersect the first 39degree line. - This will give us the approximate radius of turn. According to the chartJ the approximate radius of turn is 1599 feet. - If we continue to follow the line up to the second 36- degree line, this gives us the approximate rate of turn. The approximate rate of turn is 6 degrees per second. . Lkglet': um sewn: Method 2: Constant altitude -1 , Hank snub, 5m turn performance charts. :: {iii gure 11.14. Constant altitude turn performance. \fCenter of Gravity In an aircraft, the center of gravity (CG) is the point at which the aircraft would balance were it possible to suspend it at that point. As the location of the center of gravity affects the stability of the aircraft, it must fall within specified limits that are established by the aircraft manufacturer. The datum is an imaginary vertical plane or line from which all measurements of arm are taken. The datum is established by the manufacturer. Once the datum has been selected, all moment arms and the location of CG range are measured from this point. The Datum on the Cessna 172 is at the front face of the firewall. The arm is the horizontal distance from the 6 Cent . e reference datum to the center of gravity (CG) of an r 0f gfawty item. A moment is the product of the weight of an item multiplied by its arm. '9' (\\ Formulas Weight X Arm = Moment Total Moment / Total Weight Center of Gravity (CG) Max Ramp weight Zero Fuel weight = Usable Fuel Weight Fuel Weight /6 = Fuel Gallons lGGLL fuel weighs 6 lbs/gal Weight and Balance 1. Starting with the basic empty weight and moment we divide the moment by the basic empty weight to get the arm (line 1 armJ 39.4). 66153.1/1679.7:39.4 2. Enter the weight 'For the front pilots, rear passengers and baggage. By adding all weight together, we get the zerofuel weight. 3, Then to get the arm For line 1-4 we reference the POH For the published stations and arm. Fuel is at a published 48 CG 4, Now we enter our fuel weight, we convert fuel to lbs. fuel gallons*6:fuel weight (lbs). 5. Subtract taxi fuel and fuel burn to get the Take off weight and landing weight 6. Get the CG for all remaining lines by dividing the moment by the weight. 7. Plot the Take off weight and CG, as well as the landing weight and CG on the manufacturer's envelope (next page) in} Bas'c Empty Waight Front Pilots Rear Passengers Baggage 120le (max) Zero Fuel Weight Usable fuel Ramp Weight Taxi Fuel Takeoff Weight Fuel Burn Landing Weight Weight & Balance Cessna 172 Weight _ CENTER-OF-GRAVITY LIMITS B1670 Airplane C.G. Location - Millimeters Aft of Datum (FS 0.0) 875 925 975 1025 1075 1125 1175 1225 900 | 950 1000 1050 1100 1150 1200 2600 1150 CG Envelope 2500 Maximum Takeoff Weight Center-of-Gravity 2550 Pounds Limits 1100 2400 1050 2300 2200 1000 2100 Normal Category 950 Take off weight 2564.3, 43.7 CG Utility Landing weight 2444.3, 43.5 CG Loaded Airplane Weight (Pounds) 2000 Category 900 Loaded Airplane Weight (Kilograms) Plotting the weight and CG 1900 850 We are outside the envelope 1800 800 1700 750 1600 700 1500 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 Airplane C.G. Location - Inches Aft of Datum (FS 0.0)Moving weight and Weight & Balance Cessna 172 loading less fuel to Weight X Arm = Moment Basic Empty Weight 1679.7 39.4 66153.1 Front Pilots 345 37 12765 land in the envelope. Rear Passengers 190 73 13870 Baggage 120lbs (max) 40 95 3800 Zero Fuel Weight 2254.7 42.8 96588.1 . If we move the front 150 Usable fuel 192 48 15264 Ibs passenger to the rear Ramp Weight 2446.7 111852.1 seat and move the 1701bs Taxi Fuel 8.4 48 403.2 passenger into the front Takeoff Weight 2438.3 45.7 111448.9 seat, as well as remove Fuel Burn 120 48 5760 some fuel, we now changed Landing Weight 2318.3 45.6 105688.9 our CG and over-all weight to fall into the safety parameters of the manufacturers CG envelope.CENTER-OF-GRAVITY LIMITS 84070 Airplane C.G. Location - Millimeters Aft of Datum (FS 0.0) 875 925 975 1025 1075 1125 1175 1225 900 950 1000 1050 1100 1150 1200 2600 1150 2500 Maximum Takeoff Weight Center-of-Gravity 2550 Pounds Limits Plotting the weight 1100 2400 1050 2300 / CG change 2200 1000 2100 Normal Category 950 . Take off Weight 2438. 31bs, 45.7 CG Utility Loaded Airplane Weight (Pounds) 2000 Category 900 Loaded Airplane Weight (Kilograms) 1900 . Landing Weight 2318. 31bs, 45.6 CG 850 1800 800 1700 750 1600 700 1500 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 Airplane C.G. Location - Inches Aft of Datum (FS 0.0)