Cessna Skyhawk C-172, by Textron Aviation

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by Michael Kennedy0 views·Updated Mar 31, 2022

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  1. Introducing the Cessna Skyhawk C172 by Textron Aviation.
  2. This is part of a S. C I 309 aerodynamics with Butch Bennett for the March 2022 term.
  3. This particular module will look at the aircraft selection, airfoils,
  4. research and the physical laws associated with computing.
  5. Airfoil performance. Let's introduce the C 1 72 Skyhawk.
  6. This is a four seat single engine aircraft traditionally used for training here
  7. in the United States and has been in continuous production since 1948.
  8. This particular Skyhawk offers a very simplistic fight flight characteristics
  9. including now updated avionics, including the G 1000 series,
  10. lower landing speeds,
  11. lenient stall speeds,
  12. which will address through some performance calculations later and
  13. they're the ideal aerodynamics for a training aircraft.
  14. Today.
  15. The general aircraft characteristics are very basic. It is a four seat aircraft.
  16. One crew required.
  17. The length is relatively small at 27 ft two inches
  18. with a wingspan of 174 square feet for the wing area
  19. aspect ratio is 732. Very important to this research is the airfoil type.
  20. The knack of 24 12 with our empty weight being £1680 although in
  21. most of our calculations will be using a max gross weight of £2550.
  22. Making this our Maxwell useful load
  23. of £878 and a max payload of £870.
  24. The performance of the aircraft is relatively straightforward maximum cruise.
  25. About a 124 max cruise speed in a trainer.
  26. Let's go ahead now and jump into some of
  27. the resource requirements for airfoil and the behavior.
  28. The main airfoil data shows this to
  29. be a relatively straightforward training aircraft.
  30. The airfoil tools were used to be able to define the cord,
  31. the camera and the upper and lower surfaces of the 24-12 airfoil service.
  32. Let's look into the angle of attack.
  33. Now we selected to generate a data model.
  34. We used Microsoft sequel server to go ahead and
  35. load in all the naka data across all aircraft
  36. types so that we could look at the stall AOA A the CL levels by Naca type.
  37. When we looked at the max C. L. And the C. L.
  38. For minimum here to evaluate the AOA and Cl for the Skyhawk.
  39. We came out with the stall A. O. A at 16.5 degrees.
  40. This would be the critical angle of attack in which an aircraft
  41. would stall again and will reference this in our performance calculations later.
  42. This is a respective of weight
  43. in speed
  44. and altitude
  45. on the critical angle of attack which is at the C. L.
  46. Max of 1.582 and r zero lift A OA is roughly a negative 2.250 degrees.
  47. Now let's look at some airfoil comparisons.
  48. Not part of the assignment but I thought this might be keen
  49. to look at in terms of assessment products as a whole.
  50. As we start to define.
  51. Looking at the airfoils and the AOA
  52. characteristics around the airfoils for Cessna products.
  53. But I wanted to bring this one up relatively quick to show the difference between
  54. transitioning pilots for a jet versus those who come from the piston 1 72 world.
  55. You can see the overlay of that airfoil with respect to chord line.
  56. In Canberra is almost identical.
  57. We went ahead and used the data model
  58. that we had selected previously which the knack to type here for
  59. the citation one Cessna product is the 2301 to knock a type
  60. but you'll see that the stall, the C. L.
  61. Max and the zero lip Aowei characteristics are almost identical to a Cessna 1
  62. 72 with remarkably small differences noted in red between the stall the C.
  63. L. Max and the zero lift a away.
  64. This provides trainers the ability to
  65. move from training aircraft within the Cessna
  66. suite of products to aerodynamically similar aircraft
  67. here represented in their jet sweets.
  68. Now,
  69. as the requirements of this assignment requires
  70. us to look at the symmetrical airfoil,
  71. I went ahead and loaded a data model again for the 24 12 Naca type on the Cessna 1 72
  72. and the very first Naca type for a symmetrical airfoil,
  73. you'll see the airfoil down there below in red which is the 24 12 Cessna 1
  74. 72 and the purple which is the F 1 11 or 006 symmetrical airfoil type.
  75. You can see in the data model that the stall AOA A. The C. L.
  76. Max and men are very different based on the type
  77. of naka selected in the data model.
  78. There are some rough differences in terms
  79. of the symmetrical airfoil which distinguishes itself different
  80. from what we see in the 24-12 with the identical upper and lower services.
  81. You can notice that here in the lower. Reference in purple
  82. there is no zero lift a away due to the
  83. cord line and the mean camera line being identical.
  84. If you look at the data model to the left and reference the blue,
  85. excuse me, the purple arrows, you'll see the very bottom one where the zero A.
  86. Oa indicates zero lift.
  87. Now the non symmetrical to 4 12 naka provides
  88. greater lift efficiency at any given a o.
  89. A.
  90. With a better lift to drag ratio,
  91. with improved stall characteristics for a non symmetrical
  92. airfoil wing.
  93. Alright, let's continue on and look at some of the calculated behaviors
  94. that we have for the Cessna 1 72 aircraft.
  95. Many of the performance and limitations of the
  96. skyhawk are provided within the flight operations,
  97. manual of the pilot operating handbook but we're going to confirm
  98. some of those performance factors by
  99. looking at specific calculations that address
  100. speed,
  101. wait altitude and temperature to arrive at calculated stall
  102. speeds and lift deficiencies for the Cessna 1 72
  103. here in the calculated stall speed.
  104. We're gonna go ahead and address this from a gross weight perspective of £2550.
  105. The wing area. The Cessna 1 72 is 174 square feet.
  106. And in this demonstration we're going to look at the calculated stall speed
  107. At c. l.
  108. max at 1.58 To this would be the 16.5 critical angle of attack for this aircraft.
  109. We're going to do this at an altitude of
  110. 5000 MSL which gives us a density ratio of 50000.8617
  111. And the stadium.
  112. The standard data points here used at 2992.
  113. We use the calculated stall stall speed equation here which then gives
  114. us a stall speed of 56.32 indicated on our air speed indicator.
  115. Now we're going to reference the calculated stall speed later when we start to look
  116. at the cl differences between cl cl max and how those might exceed C.
  117. L. Max
  118. and what contributions they have to stall speed.
  119. In a later slide
  120. Let's go ahead and jump into the lift efficiency.
  121. We have four conditions in which were required to examine lift deficiency
  122. for this particular aircraft and demonstrate the capabilities of the Cessna 172.
  123. Condition one looks at how lift changes with their speed.
  124. You have a constant AO AO AO and altitude
  125. are held condition to how lift changes with altitude.
  126. If the constant AOA and airspeed are held.
  127. Whereas condition three is,
  128. how do the required cln ao a for the Skyhawk change with changes in air speed.
  129. And finally what happens if the required cl is larger than the C. L.
  130. Max of the Skyhawk airfoil which we referenced just a
  131. minute ago as we looked at the stall speed.
  132. Let's look at condition one here.
  133. Condition one requires us to look at the change with their speed.
  134. If a constant AOA announced that are held with respect to lift
  135. here,
  136. we're going to have the skyhawk traveling initially at 55
  137. knots and then have its speed up to 110 knots.
  138. For this particular example,
  139. I am using max cl cl max at the 1.582 which would be that 16.5 degree angle of attack.
  140. The critical angle of attack.
  141. I'm doing this for two purposes to demonstrate here uh that the aircraft is
  142. pretty much at its full weight value at the 55 knots at £2432.29.
  143. But when we take that aircraft still at 5000 ft to 110 indicated knots,
  144. it's 9720 nine's of pressure to try to hold that aircraft at 110 knots
  145. at 16.5 degrees which this aircraft won't be able to do.
  146. So this begs the question, can this aircraft handle that?
  147. Certainly at the 55 knots at 5000 ft,
  148. the pounds of lift is appropriate for this airfoil.
  149. However, when you look at this type of category of an aircraft,
  150. being a normal category rather than a utility
  151. category of the aircraft,
  152. we have a part 23 limitation to 3.8 Gs load limit on the airframe,
  153. which would bring this aircraft's weight load limit at 3.8 to £9690
  154. in which the 110 knots at 5000 MSL would exceed that. So, with £39 over,
  155. this shows us that we need to keep
  156. this aircraft within a limited speed factor to avoid
  157. structural damage but to stay inside the stall envelope for this given airframe,
  158. let's look at condition number two,
  159. condition number two is how does lift change with altitude? If a constant A. O.
  160. A and airspeeder Hill.
  161. So similar to condition one will test
  162. the relationship between lift as altitude increases.
  163. We should note that there is a direct relationship
  164. between altitude and density ratio lift will decrease,
  165. the pounds of lift will decrease as altitude increases.
  166. So you can see first
  167. At 100 knots,
  168. this aircraft will be at 1000 ft MSL and will then
  169. climb this aircraft to 5000 ft. MSL and you can see that the £2131
  170. will be at the 1000 MSL rate, but as we increase altitude,
  171. the lift will decrease and the formula here for that proves as we've
  172. seen about a £200 decrease between the 404,000 ft that we increased in.
  173. MSL
  174. finally, let's look at number three.
  175. Number three is, how do the required cl in the airway for the sky up,
  176. Change with changes in air speed.
  177. And we've done this several ways in which the sky hop will initially travel
  178. at 100 K. I. S indicated air speed and then reduce the speed to 75 knots
  179. At the gross weight of 2550 pounds.
  180. Again using the same given parameters 174 square feet,
  181. it will then travel at 3000 ft. MSL with the appropriate density ratio.
  182. So you can see with the formula that we've used here
  183. that the 30000.472 demonstrated at full weight at 100 knots,
  184. But in order to achieve the same altitude at a lower air speed.
  185. In the second formula at 75 knots, it's 0.839,
  186. we would have to increase the back pressure and increase
  187. our angle of attack to maintain that same altitude.
  188. So, at a lower airspeed,
  189. you're required to increase the AOA to maintain that same altitude.
  190. The cl must increase at the lower speed to maintain the same altitude as well.
  191. Some AOA and clb match there.
  192. Mhm.
  193. Now the example that we give at the bottom is cautionary as you increase the weight,
  194. you will stall at a higher indicated airspeed,
  195. the stall at the speed of which an
  196. aircraft stalls is not influenced directly by speed,
  197. weight or temperature.
  198. In terms of the
  199. angle in which the aircraft stalls. A critical angle of attack is the only component
  200. in terms of the
  201. lift or the airfoil losing all lifting properties.
  202. But you will arrive at that critical angle of
  203. attack at different speeds based on your weight.
  204. And we're looking at the differences here between
  205. the £1900 aircraft and a 25 £150 aircraft.
  206. You can see it increases by seven knots.
  207. You will arrive at the stall speed at the max cl at a higher airspeed
  208. than you would at a lower airspeed at a lower weight of the aircraft.
  209. And finally what happens if the required cl is larger than the C. L.
  210. Max of your of your airfoil and what
  211. speed regime is usually associated with that condition.
  212. And again what we're doing here is demonstrating the
  213. maximum performance of the aircraft in terms of C.
  214. L. Max and the 16.5
  215. degree A. O. A. The critical angle of attack
  216. and what speed you would arrive at that at max gross weight which here is 56.32 knots.
  217. So in this example if the aircraft CLS assumed it's a 1.61
  218. that would exceed that which is of 1.58 to the C.
  219. L. Max of this aircraft here demonstrated at 5000 ft. MSL
  220. which would mean that the airfoil would be stalled.
  221. The equation above demonstrate that the skyhawk at gross weight at 5000 MSL
  222. would reach the critical angle attack at that indicated air speed at 56.32.
  223. So based on the condition that we used in three
  224. pilots must fly with a higher AOA at lower speeds to maintain the same altitude
  225. or to hire a away at lower speeds to avoid a stall but not to the point in
  226. which they reached the critical angle of attack during
  227. the increased a a way to maintain altitude.
  228. I hope you've enjoyed this presentation.
  229. This has been my approach to aircraft selection and the airfoils,
  230. research and physical laws associated with calculating
  231. behaviors of the Cessna 1 72.
  232. Thank you.