   Air plane and Helicopter
  Just wondering if anyone had any good tutorials on air plane or helicopter physics. I have some ideas about going about it but I would prefer to go the with using the actual physical formulas for calculating lift, self leveling, and such and then optimize.Thanks for any help. I will also be scanning the web and post up here if I find anything. 
  self-leveling? what do you mean? ;  check out tbe book Physics for Game Developers ;  I was wondering about how helicopters seem to self-level. But for now I read up on plane physics.AirplaneSeems simple enough.Main formula (lift and drag nearly identical)L = 1/2 * d * V^2 * S * ClD = 1/2 * d * V^2 * A * Dcd=air densityV=veloctiyS,A = area of wing, or area of rigib bodyCl, Dc = coefficient of lift, coefficient of dragI think drag is right I might have to double check... but basically you take your velocity, square it, and mutiply it by a couple arbitrary numbers to get your lift. Same with drag.So basically once you thrust reaches you to a velocity in which your lift force is greater than your weight force, you should become airborne. You also have to worry about drag which will determine your terminal velocity just as much as thrust.So if you plane is banking hard, you appearantly are actually falling since nothing is counter-acting against your weight force.Also to get quick elevation you translate some of you thurst force in an upward direction (pulling up) which adds more to counter-act your weight force thereby getting a quick incline...now to helicopters which I am sure will not be easy...   [edited by - JimmyNelson on February 3, 2004 12:00:04 PM];quote:Original post by jimmynelson  L = 1/2 * d * V^2 * S * ClD = 1/2 * d * V^2 * A * DcYou should be consistent in your notation. Use CL for lift coefficient (capital ''L'' implies a 3-dimensional lift. A lower-case ''l'' implies a 2-dimensional lift.) And use CD for drag coefficient. Also, for airplanes, the reference area is the same reference area used for lift, so you should have (and this is only a notation edit):L = 1/2 * rho * V^2 * S * CLD = 1/2 * rho * V^2 * S * CD  rho, the greek letter rho, is the traditional symbol for air/fluid density. And, S is the reference area for both drag and lift. S typically is taken to be the total wing area projected into the horizontal plane.Now, having those two equations is almost no help at all for a simulation!! You need values for CL and CD to calculate the lift and drag force. But, what values are you going to choose for CL and CD? Lift is a function of angle attack, along with the geometry of the wing (aspect ratio, wing sweep, etc.) and ultimately the speed of flight. Lift can be approximated as a linear function for small angles of attack (less than, say, 10 degrees). Drag can be written as a parabolic function of lift, again only for small angles of attack.If you search the forum archives ("CL", "drag", "lift", etc.) you should find some old threads that give some useful formulas for approximating CL and CD.The other thing you need, which enables you to understand the self-leveling, is a pitching moment. The equation is:Pitching_Moment = 1/2 * rho * V^2 * S * chord * CMwhere CM is the pitching moment coefficient, and chord is the average chord width of the wing.For small angles of attack, pitching moment can be represented as a linear function of lift. For an airplane to be self-leveling, the slope of the CM-CL graph must be negative (pitching moment decreases as lift increases), and the CM curve must cross through CM = 0 for some achievable value of CL (on the order of -1.0 to 1.0 or more for very efficient wings). If CM is zero at a CL that corresponds to a lift force equal to the weight of the airplane, then the airplane will self-level at exactly the right lift for flying straight and level.quote:Original post by jimmynelson  So if you plane is banking hard, you appearantly are actually falling since nothing is counter-acting against your weight force.If you are banking with the wings straight up and down, you won''t be generating much lift against gravity, but you may be generating some. The fuselage can generate lift when the airplane has banked 90 degrees. But, just as easily the fuselage can generate a downward lift force that will cause you to accelerate faster than free fall. Further, the wings almost never lie in a plane, e.g., there is a dihedral angle between the wings. Because of this, when the plane is actually turning in a bank (vs. flying in a straight knife-edge line), there can be a component of lift generated by the wings that also either counteracts gravity or acts in the same direction as gravity....The real physics of banking is tricky. Your equations above don''t begin to address this.quote:Original post by jimmynelson  Also to get quick elevation you translate some of you thurst force in an upward direction (pulling up) which adds more to counter-act your weight force thereby getting a quick incline...I worked in the full scale wind tunnel at NASA Langley for a year back around 1991. At the time they were just starting to free fly the X31 thrust-vectored concept fighter jet. I wasn''t allowed to see any of the tests of this thing, but later it was publically revealed that the X31 can fly horizontally while pointing almost straight up in the air. The thrust fully balances the weight and also provides a component of force to move the airplane forward. Of course, some of these fighter jets can act like rockets, requiring no lift at all---just engine.http://www.dfrc.nasa.gov/Gallery/Photo/X-31/Small/EC94-42478-4.jpgNote that in this photo the airplane is not climbing. It is moving horizontally to the right.quote:Original post by jimmynelson  now to helicopters which I am sure will not be easy...   There are some fairly simple flight models for helicopters. But, that''s not my forte.There is a book called "Helicopter Theory" by Wayne Johnson, published by Dover. Honestly, it isn''t a great book. You will find it difficult to impossible to use. But, you may be able to find a copy cheap somewhere, since Dover basically provides cheap paper back technical books----reprints of classic texts. Graham RhodesSenior ScientistApplied Research Associates, Inc. ;quote:The thrust fully balances the weight and also provides a component of force to move the airplane forward. Of course, some of these fighter jets can act like rockets, requiring no lift at all---just engine.A production example of a plane that generates enough thrust to counter-act its weight is the Harrier jump jet (whatever it gets called in the states). Of course there are others, athough apart from the harrier they need to be in flight to make use of their excess thrust, as far as I''m aware.There''s a story that a British test pilot (working for Rolls Royce) thought it would be funny to land one without the benefit of the engine - when he did touch down he was doing a fairly silly speed & the wheel brakes aren''t desined to stop it from that speed either...the results were fairly predicable, athough he did survive. ;  Thanks grhodes_at_work,I always enjoy your posts. I was just wondering if you think for pratical game usage that I could get away with ignoring the reference area for drag and lift. I was just gonna set them someplace that works and go with it. My plane would fly and the drag and lift wouldn't be realistic per say... but I would be creating a terminal velocity, elevation change, etc... And I don't really understand jet physics so much. You would have to have some lift though, to counter-act the inevitable force of gravity (even bullets can't ingnore this.) I also never understand pictures like that, where the jet can almost be perpendiculat to the ground and climb without engine failure (man it sure would be fun to be in it though).I also have read before that although jump jets can take off in very little space, they burn up such a large amount of fuel on take-off and landing that is wasn't the most practical jet.   [edited by - JimmyNelson on February 3, 2004 9:08:01 PM];quote:Original post by jimmynelson  I also have read before that although jump jets can take off in very little space, they burn up such a large amount of fuel on take-off and landing that is wasn''t the most practical jet.   In any manouvre where gravity is completely countered only through engine thrust, the total energy for this manouvre must be derived from chemical potential energy... which means that while certain aircraft (Harrier/AV8,X31, etc) can counter gravity with thrust alone, they cannot do it for extended periods and maintain effective flight endurance.The Harrier was originally designed to be effective on bomb damaged runways and to be deployable from short roads, making it harder for an enemy strike to neutralise the British air defences as often happened during the early days of WW2.For the original poster... you need to make another decision with your aircraft sim: are you going to utilise a point-mass approximation for your aircraft, or are you going to simulated the lift and drag components of airframe parts (wings, tailplane, fuselage, undercarriage, etc.)? While the former is far easier to implement, it is also far less realistic.Good luck with your project,Timkin ;  Doesn''t lift depend on the angle of attack? As does drag. I don''t see it in the formulas? ;/   ---shurcooL`;  Timkin:I am actually using Tokamak''s Rigib Body simulator. I am creating just a simple point mass (rigib body) and it will be used in a large mutiplayer type game (FPS not flightsim). I am wanting to add as little math as possible, seeing as there might be 6-7 of these planes in the world as well as many other objects.shurcool:It is in there... L = 1/2 * rho * V^2 * S * CLD = 1/2 * rho * V^2 * S * CDS = is the reference area for both drag and liftWhy parachutes work so well, S jumps up from nearly nothing too creating a terminal velocity around 5 m/s and reaching that in approx 3 meters... so says my physics book... 