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Flight formulas

  1. Mar 24, 2009 #1
    Hello !

    I'm a Computer Science student and I'm working on a flight simulator. I know that there's a Computer Science section but my question is more about flight formulas. I have good bases in mathematics but in physics, it's completely different. I would like to know what formulas affect planes in flight and which forces affect planes in flight.

    Thank you. :smile:

    Sorry if my English is not excellent, it's my second language.
     
  2. jcsd
  3. Mar 25, 2009 #2
    Hi

    I'm a new in this forum too.

    I'm an old flight-sim fart who started own 'open source flight-sim' hobby project basically just to refresh the math & physics skills, before the the “Dad! Please, could you help me a bit with THIS?!” time is in the hand.
    ”Ehh.... Ohhh... Sure...Son.”


    Vok5,
    You can find a lot about the basics and whole lot more with google by yourself.

    You will find a many pages about the 'Basic four forces: Lift, Weight, Thrust & Drag' but this is simplifying the problem of the flying way too much (imho). These forces can calculate nicely individually, but you wont get a point of flying ONLY from those.

    You must add the Torque/Moment element to be able to fly in controllable manner:
    http://www.google.fi/search?hl=fi&c...=result&cd=1&q=longitudinal+stability&spell=1

    The forum search seems to also give many hits with 'aerodynamics' word:
    https://www.physicsforums.com/search.php?searchid=1524259 [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Mar 25, 2009 #3
    Thank you for your answer.

    I searched on Google but I didn't really understand formulas that I found. I'll dig more deeply about Torque/Moment element. With some help from my teacher, I'll probably be able to create something good :smile:.
     
  5. Mar 26, 2009 #4
    Drag=1/2(v^2)(A)(C)(P)
    p=air denstiy
    A=cross sectional area
    C= drag coeffiecient anywhere from (.25)-(.45)
    v=velocity
     
  6. Mar 27, 2009 #5
    Hi Vok5

    The traditional designed airplane moment principal is like old fashion hook counter-weight balance measure.
    The plane must be able to carry the support point from where it's hanging.

    Just an idea picture
    http://fdm4bge.1g.fi/Files/10001/apics/ac_moments.jpg [Broken]

    The 'Wing Lift' must counter the 'Plane weight + Tail Lift' load, but same time the 'Plane weight' caused moment must be countered by 'Tail Lift'*tail_arm moment load, for the level flight.
     
    Last edited by a moderator: May 4, 2017
  7. Mar 27, 2009 #6

    rcgldr

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    The center of gravity is normally in front of the center of lift (or else the aircraft is computer assisted) for stability. In the diagram above, the plane would pitch up unless the tail were generating positive (+) lift.
     
  8. Mar 28, 2009 #7
    Hi

    I assume you are preferring with center of lift a wing center of lift only. This is incorrect, because we are looking how traditional airplane wing and tail are working together.

    The Lift is acting from Neutral Point (NP) and that's the reason why plane in diagram is flying with negative tail lift.

    More about Stability Concepts
    http://adamone.rchomepage.com/index5.htm [Broken]

    Aircraft Center of Gravity Calculator
    http://adamone.rchomepage.com/cg_calc.htm [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Mar 28, 2009 #8

    rcgldr

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    Homework Helper

    I meant the center of lift for the entire aircraft. The first web site you linked to above includes this picture:

    np.gif

    The neutral point is the where the center of gravity needs to be located so that an aircraft has zero pitch stability (not positive or negative). The center of lift can be considered the vector sum of all aerodynamic forces on an aircraft. For a symmetrical airfoil (no pitching moment), they're effectively the same. If the airfoil is cambered, then they are slightly apart; the neutral point is further back for a wing with positive camber because of the pitch down aerodynamic torque.

    You're diagram is a bit confusing because it appears to depict a symmetrical airfoil, no pitching moment (aerodynamic torque), and the center of lift for the entire aircraft and neutral point should be the same, so it appears you show the neutral point too far aft of where it should be. Without a pitching moment, then with the center of gravity and the tail's negative lift both being behind the center of lift for the main wing, the result should be a pitch up (clockwise) torque.

    As mentioned above, you have linear forces (lift, weight, thrust, drag) and angular torques (pitch, roll, yaw). The comment about moment above was about pitch stability, which occurs when an aircraft self corrects to level flight at a specific speed without any control movments. The math of aerodynamics is very complex. There are programs, like xfoil, that can calculate airfoil parameters. How complicated do you want your flight simulator to get? The best consumer flight simulators seem to be the radio control model simulators, such as RealFlight. I don't know how accurate the modelling is in programs like FSX.
     
    Last edited: Mar 28, 2009
  10. Mar 28, 2009 #9
    In practice the CG must be front of NP(lift), just like in your linked picture left side plane.

    “For an aircraft to be stable in pitch, its CG must be forward of the Neutral Point NP by a safety factor called the Static Margin, which is a percentage of the MAC (Mean Aerodynamic Chord). Static Margin should be between 5% and 15% for a good stability.”

    The professional pilots need to pay a lot of attention to ensure the CG location will stay within specification, while loading the plane with fuel, cargo & passengers.

    The diagram is simplified in purpose, because the idea is to present how the tail force is countering the weight caused nose down moment. The use of symmetrical or asymmetrical airfoil wont change this concept.
     
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