1 / 30

Aeronautical Engineering for Pilots

Aeronautical Engineering for Pilots. (1643 -1727) Or: how I became a fan of Isaac Newton Or: Applied physics for fun and profit. Newton’s laws of motion are the basis for most of classical aerodynamics, especially the 2 nd law. F= M*A. Force equals mass times acceleration.

judithdaniels
Download Presentation

Aeronautical Engineering for Pilots

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Aeronautical Engineering for Pilots (1643 -1727) Or: how I became a fan of Isaac Newton Or: Applied physics for fun and profit

  2. Newton’s laws of motion are the basis for most of classical aerodynamics, especially the 2nd law F= M*A Force equals mass times acceleration With a little algebra we can rearrange the 2nd law to read: F= _M_ * Velocity second

  3. Mr. Newton presiding in Trinity Chapel Cambridge, England

  4. Let’s use Mr. Newton’s laws to analyze the Boston Tea Party Samuel Adams looks like he’s throwing a tea chest at about 10 MPH. Tea’s light but the chest is mahogany so maybe 32.17 pounds??? Sammy imparts a horizontal velocity of 10 mph on tea chest in 1 second Throwing Tyranny overboard: Boston Tea Party reenactment

  5. Story Problem # 1 Release Start Dist. Time = 1 sec Time = 0 Velocity = 14.7 FPS Velocity = 0

  6. Sammy pushes on boat with the same force as on the tea chest, 14.7 pounds. (Newton’s third law)

  7. Hainan Island China, 2008 So now let’s apply the same 2nd law to this Learjet Model 36 ER engine but use air molecules in place of tea chests. The Garrett TFE 731 “swallows” 115 lb. / second of air molecules at full thrust. Exhaust velocity at the engine exit has to be higher than the airplane’s top speed MMO=.81 460 Kt. 776 FPS Exhaust Velocity ~976 FPS Force (Thrust)= 115 # per sec / 32.17 * 976 FPS = 3488 #

  8. A couple of guys that knew a lot about AIR! "I am convinced that human flight is both possible and practical."Wilbur Wright, 1899

  9. The Lady that helped them become famous

  10. What will humans think of these photos in the year 2519?

  11. Airfoil Theory Wright kite at Kitty Hawk

  12. Equal molecule travel time ------------------------------------------------------------------------------------------------- Here’s something like the Wright’s airfoils Distances are equal, top & bottom????

  13. Cessna 172: Gross weight – 2300 Pounds Span – 36 feet V best glide ~ Vy – 80 MPH, 69 Kts. Best L/D α – 4 degrees

  14. Cessna 36’ foot span wing flying in streamtube 4 degrees 36 Feet

  15. Downwash WeightLift α Relative Wind

  16. Drag due to lift Newton’s 2nd: Force= M /sec * Velocity Flight path angle + α WeightLift α Relative Wind 0 Downwash Horizontal

  17. At L/D max Parasite drag and induced drag are about equal (windmilling prop) Total drag = 2 * Induced drag = 2 * 324 pounds = 648 pounds Required HP = Drag * velocity = 648 lb. * 117 Fps / 550 = 138 Horsepower Available HP = Engine HP * Propeller efficiency = 160 HP * 85% = 136 HP

  18. Aircraft Structures 1952 Northrop F-89 Scorpion Wing Failure

  19. FAR’s structural certification standards – Small aircraft §23.2230 Limit and ultimate loads. The applicant must determine— (a) The limit loads, which are equal to the structural design loads unless otherwise specified elsewhere in this part; and (b) The ultimate loads, which are equal to the limit loads multiplied by a 1.5 factor of safety unless otherwise specified elsewhere in this part. §23.2235 Structural strength. The structure must support: (a) Limit loads without— (1) Interference with the safe operation of the airplane; and (2) Detrimental permanent deformation. (b) Ultimate loads. (can come back bent)

  20. What are moments and what are they used for? Calculate moments about this point F1 F2 40 Lb. 3’ 6’ 6’ Girl moment = 40 lb. * 6’ = 240 foot – pounds To balance (moments equal): F1 = 240 foot – pounds / 6’ = 40 pounds F2= 240 foot – pounds / 3’ = 80 pounds

  21. 2’ 4.5’ 7.5’ 6’ F1

  22. What is a spar cap and why’s it important?

  23. Material Stress Analysis Wing spar outboard of strut attach, Length = 9.5’ Lift at ½ span from strut attach Moment Tip Strut Attach 8 in.

  24. Moment = 3,339 ft. – lb. Compression Tension 8”=.66’ Immovable object Spar

  25. Approximate 2024 Aluminum Stress Strain curve Proportional limit .002 in. / in.

  26. Thank you, Terry Spath terryspath@yahoo.com 503-812-0606

More Related