Aerodynamic Center1 Suppose We Have the Forces and Moments

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Aerodynamic Center1 Suppose we have the forces and moments specified about some reference location for the aircraft, and we want to restate them about some new origin. N ~ L N ~ L M AAM z ref new x x ref xnew M ref = Pitching moment about xref M new = Pitching moment about xnew xref = Original reference location xnew = New origin N = Normal force ≈ L for small ∝ A = Axial force ≈ D for small ∝ Assuming there is no change in the z location of the two points: MxxLMref=−() new − ref + new Or, in coefficient form: xx− new ref CCCmLM=− + ref cN new mean ac.. The Aerodynamic Center is defined as that location xac about which the pitching moment doesn’t change with angle of attack. How do we find it? 1 Anderson 1.6 & 4.3 Aerodynamic Center Let xnew= x ac Using above: ()xx− CCC=−ac ref + M refc LM ac Differential with respect to α : ∂CM xx− ∂C ref ac ref ∂CL M ac =− + ∂∂∂αααc By definition: ∂CM ac = 0 ∂α Solving for the above ∂C xxac− ref M ref ∂C =−L , or c ∂∂αα ∂C xxac− ref x ref M ref =− ccC∂ L Example: Consider our AVL calculations for the F-16C • xref = 0 - Moment given about LE ∂C • Compute M ref for small range of angle of attack by numerical ∂CL differences. I picked αα= −=300 to 3. x • This gave ac ≈ 2.89 . c x ∂C • Plotting Cvsα about ac shows M ≈ 0 . M c ∂α ∂C Note that according to the AVL predictions, not only is M ≈=0 @ x 2.89 , but ∂α ac also that CM = 0 . The location about which CM = 0 is called the center of pressure. 16.100 2002 2 Aerodynamic Center Center of pressure is that location where the resultant forces act and about which the aerodynamic moment is zero. Changing “new” to “cp” and “ref” to “NOSE” to correspond to AVL: xxcp − NOSE CCC=−+ MLMNOSE cp c x C ⇒=−cp M NOSE cCL So if: CC∂ MMNOSE= NOSE , CCL ∂ L this will be true. This means that CCML≈=0 at 0 . 16.100 2002 3 Cm vs Alpha for F16C from AVL (M=0) 1.5 1 0.5 0 -10 -5 0 5 10 15 20 -0.5 Xref = 0 -1 Cm Xref = Xac = 2.89 -1.5 -2 -2.5 -3 -3.5 Alpha Wind Tunnel Test Aerodynamic Center Characteristics for Washout and Rigid Wings (Altitude = 10,000 ft.) 50 45 40 35 Rigid Wing Washout Wing 30 A.C. (% MAC) 25 20 15 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 Mach Number.

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Equations for the Application of to Pitching Moments
https://ntrs.nasa.gov/search.jsp?R=19670002485 2020-03-24T02:43:10+00:00Z View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by NASA Technical Reports Server NASA TECHNICAL NOTE NASA TN D-3738 00 M h U GPO PRICE $ c/l U z CFSTl PRICE@) $ ,2biJ I 1 Hard copy (HC) ' (THRU) :: N67(ACCES~ION 11814NUMBER) 5P Microfiche (MF) > Jo(PAGES1 k i ff 853 July 85 2 (NASA CR OR TMX OR AD NUMBER) '- ! EQUATIONS FOR THE APPLICATION OF * WIND-TUNNEL WALL CORRECTIONS TO PITCHING MOMENTS CAUSED BY THE TAIL OF AN AIRCRAFT MODEL by Harry He Heyson Langley Research Center Langley Stdon, Hampton, Vk NATIONAL AERONAUTICS AND SPACE ADMINISTRATION WASHINGTON, D. C. NOVEMBER 1966 . NASA TN D-3738 EQUATIONS FOR THE APPLICATION OF WIND-TUNNEL WALL CORRECTIONS TO PITCHING MOMENTS CAUSED BY THE TAIL OF AN AIRCRAFT MODEL By Harry H. Heyson Langley Research Center Langley Station, Hampton, Va. NATIONAL AERONAUTICS AND SPACE ADMINISTRATION For sale by the Clearinghouse for Federal Scientific and Technical Information Springfield, Virginia 22151 - Price 61.00 , EQUATIONS FOR THE APPLICATION OF WIND-TUNNEL WALL CORmCTIONS TO PITCHING MOMENTS CAUSED BY THE TAIL OF AN AIRCRAFT MODEL By Harry H. Heyson Langley Research Center SUMMARY Equations are derived for the application of wall corrections to pitching moments due to the tail in two different manners. The first system requires only an alteration in the observed pitching moment; however, its application requires a knowledge of a number of quantities not measured in the usual wind- tunnel tests, as well as assumptions of incompressible flow, linear lift curves, and no stall.
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