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Airplane loop - help please

  1. Aug 17, 2014 #1
    1. The problem statement, all variables and given/known data
    I have to write a C# program for an assignment for school. Programming is not a problem, but physics is :P So I am kindly asking you guys for helping me out with this.. Sorry for my bad english, I'm not a native speaker.

    A glider pilot starts a back-loop with a plane, flying horizontally at a speed of 60 m/s. Maneuver is carried out so that he pulls the stick back and holds it in the same position througout the whole loop. The initial centripetal acceleration is 30 m/s2 in the direction vertically upwards.



    2. Relevant equations
    I must calculate the position, velocity and acceleration of the aircraft as a function of time during the interval from the beginning
    implementation of acrobatics to the time in which the aircraft at a constant acceleration is outlining a full circle.
    Does the aircraft come back to the original/initial position?
    The origin is placed at the point at which the airplane moves from the horizontal position of the ribbon/loop.

    I really hope I translated it so you can understand.


    3. The attempt at a solution
    I'm not really good at physics so I don't even know how to start :/

    Acceleration due to gravity is 9.8 m/s2
    Air resistance can be neglected, so that the only force acting on the plane is the force of gravity and the lift force, which is perpendicular to the direction of the current speed of the airplane.
    Surrounding air is stationary relative to Earth and a plane (throughout the maneuver) doesn't rotate around its horizontal or vertical axis.
     
  2. jcsd
  3. Aug 17, 2014 #2
    Shall we assume that the pilot maintains constant airspeed throughout the maneuver? If not, I am afraid the problem is underspecified.

    Anyway. You start with Newton's law that relates acceleration with force. I am sure you have heard about it.
     
  4. Aug 17, 2014 #3
    I forgot to add a hint which was given, sorry:

    From the initial centripetal acceleration reduced by the gravitational acceleration and velocity, we can calculate proportionality coefficient between the lift force and the square of the speed of the aircraft, which was assumed to be a constant.

    But I just don't understand it :/
    Newton's law? You mean F = m x a ?

    Thank you for a reply
     
  5. Aug 17, 2014 #4
    Due to the centripetal acceleration a plane starts to rise and the airplane due to the force of gravity loses speed
    (Because the weight is no longer working at right angles to the direction of velocity). Due to the reduced speed decreases buoyancy (as it is proportional to the square of the speed), which affects the smaller centripetal acceleration. If the speed does not decrease, to the movement of the arc upward centripetal acceleration grow as the force of gravity is no longer directed opposite to the force of buoyancy.
    The situations described are valid until the aircraft reaches the upper point of tangles, then the situation
    turns.
     
  6. Aug 17, 2014 #5
    Yes, I mean that law.

    And the hint is valuable, too. I suggest you refresh your memory on the lift force and its relationship with the square of the speed.
     
  7. Aug 18, 2014 #6
    Voko, I still can't figure it out? Do you know how to solve this? Can you please help me with direct info? I must get this right till wednesdy and still have to do the programming.
     
  8. Aug 18, 2014 #7
    I cannot solve this for you, this is against the rules of the forum. I can only provide some help and guidance.

    Newton's second law is a system of second-order differential equations, which you will have to integrate numerically. Do that sound like something familiar to you?
     
  9. Aug 18, 2014 #8
    Well, this class is called 'Numerical methods' :) It is familiar, but I suck at it :/
    Can you give me your email?
     
  10. Aug 18, 2014 #9

    HallsofIvy

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    Science Advisor

    If the pilot is, indeed, flying in a perfect circle then his acceleration must be constant and always pointing toward the center of the circle. You say "initial centripetal acceleration is 30 m/s2 in the direction vertically upwards" so his acceleration is always 30 m/s^2 pointing toward the center of the circle.

    Any circle can be given in parametric equations as [itex]x= -R sin(\theta)[/itex], [itex]y= R cos(\theta)[/itex] where, here, x is the horizontal distance right or left of the center of the circle and y is the vertical distance above or below the center of the circle. R is the radius of the circle and (x, y), when [itex]\theta= 0[/itex], is the starting position at the bottom of the loop. Assuming a constant angular speed, we can write [itex]\theta= \omega t[/itex] where [itex]\omega[/itex] is the angular velocity and t the time.

    Differentiating, [itex]dx/dt= -R\omega cos(\omega t)[/itex] and [itex]dy/dt= -R\omega sin(\omega t)[/itex].
    We are told that, at the beginning of the loop, when t= 0, the airplane has a horizontal speed of 60 m/ so we have [itex]dx/dt(0)= -R\omega= 44[/itex].

    The net acceleration is the second derivative, [itex]d^2x/dt^2= R\omega^2 sin(\omega t)[/itex] and [itex]d^y/dt^2= R\omega^2 cos(\omega t)[/itex]. We are told that, at the beginning of the loop, when t= 0, the airplane has an initial vertical acceleration of 30 m/s^2 so we must have [itex]R\omega^2= 30[/itex].

    From those two equations, [itex]R\omega= 60[/itex] and [itex]R\omega^2= 30[/itex], you can solve for both R and [itex]\omega[/itex] and so find the parametric equations for the airplane's trajectory. Add R to the y component to move the origin to the bottom of the loop.
     
    Last edited: Aug 18, 2014
  11. Aug 18, 2014 #10
    OK, thank you very much! I really appreciatte it!
    I've got another theory, so we can compare:

    Fact that this is a glider. And ignoring air drag really simplifies things. With no drag the principle of conservation of energy says that the glider will complete a closed loop, returning to its starting position and speed at the end of the loop.

    For your computer program I suggest that you divide time into small increments - say 1/10 second, and determine the change in acceleration, velocity, and position for each increment of time. The equations you will need are:

    Since lift force is proportional to the square of the velocity, so is centripetal acceleration. From a = v^2/r, this means that Kv^2 = v^2/r, and r is therefore a constant. Not what I would have expected.

    3. For each increment of time you have a starting position, velocity, and acceleration, in both x and y coordinates. You need to treat velocity and acceleration as vectors (i.e. they have magnitudes in both the x and y directions) The ending acceleration position can be calculated from

    p_2 = p_1 + v_1t + (1/2)a_1(delta_t)^2

    where v_1 and a_1 are the values of initial velocity and acceleration. Other equations:

    v_2 = v_1 - a_1 delta t

    a_2 = a_1 + a_lift - a_gravity

    where a_lift is v^2 x (30/60^2) in the direction perpendicular to the velocity vector, and a_gravity is -g in the vertical direction.

    -------------------------------------

    What do you say about what's written above?
    I have to study and try to understand both theories. It's very difficult for me.
    What does 'derivative' mean in english? Is this it -> (x2)' = 2x ?
     
  12. Aug 18, 2014 #11
    The purpose of this forum is not to help you get good marks no matter what. It is to help you learn.

    That means you must be making an effort, and you must show your effort to get help here. If you are looking for other kinds of "assistance", you will have to look elsewhere.
     
  13. Aug 18, 2014 #12
    Voko, I understand & agree with you. But what if I don't understand 'Newton's second law is a system of second-order differential equations, which you will have to integrate numerically'
     
  14. Aug 18, 2014 #13
    This is Newton's second law: $$ m \vec a = \vec F .$$ The little arrows indicate that the symbols represent vectors. In this particular case, we can assume that the aircraft moves in two dimensions only; let's call these two dimensions "distance" (left to right) and "altitude" (bottom up). The vectors are then pairs of numbers, and the law is really two equations: $$ m\ddot x = F_x ; $$ $$ m \ddot z = F_z, $$ where ##x## is the distance, ##z## is the altitude, the two dots mean "second derivative", and ##F_x## and ##F_z## are the distance and altitude components of the force acting on the aircraft. This is your system of second-order differential equations.
     
  15. Aug 18, 2014 #14
    Thank you. And this is for a glider plane right?
     
  16. Aug 18, 2014 #15
    Yes, "aircraft" = "glider". You need to understand what ##\vec F## is in this case, and find ##F_x## and ##F_z##.
     
  17. Aug 19, 2014 #16
    We are trying to solve this with Euler's method..

    I'll try to translate this correctly.. I'll mark XXX equations I'd have to fill in.


    In this case we have a plane with acceleration, which is at all times orthogonal to the movement of a plane.
    If we would describe the movement in a horizontal plane (without the effect of a gravitational force) we would get a movement along a circural path.

    1) a = a_0 x ( V_2^2 / V_0^2 )

    2) a_0


    So we have;

    V_0 ... initial speed
    XXX1 ... initial position
    XXX2 ... speed of the plane
    XXX3 ... radius of the circle
    XXX4 ... initial slope of the plane at a start

    We should get two equations:

    3) XXX5

    4) XXX6

    where XXX7 is defined as

    5) XXX8


    Here XXX9 is the speed of the plane in x direction, XXX10 is the speed of the plane in y direction.

    From this equations we get two diferential equations for the position in the x and y direction.

    6) XXX11

    and

    7) XXX12


    Somehow we have to take into account also a quadratic dependance of the components of the acceleration, perpendicular to the direction of the movement of the speed:

    a(v) + k v*v

    that's why

    a = a_0 x ( V_2^2 / V_0^2 )

    Along this aerodynamic component there is also the acceleration due to gravity, which always points down.




    Now, I don't know what XXX's are .. Anyone?


    Also, someone suggested this (also by Euler's method)

    There are four equations, because you want to calculate velocity and position in both x- and y- coordinates. Using Euler for velocity, it works:

    Vx_2 = Vx_1 +Ax t, where Ax = 39.8(V)(Vx_1)/60^2

    Vy_2 = Vy_1 + Ay t, where Ay = 39.8(V)(Vy_1)/60^2 - 9.8

    And then Euler can be used again for position:

    x_2 = x_1 + (Vx_1)t
    y_2 = y_1 + (Vy_1)t



    Still trying to figure out those XXX's
     
  18. Aug 19, 2014 #17
    If you assume no gravity (and no drag), then both the velocity and acceleration remain constant, and the glider will be flying in a circle. You do not need to integrate anything in that case. The complete solution for this case has already been given by HallsofIvy.

    If, however, gravity is present (and this is what the problem states), all of what you wrote before "Also, someone suggested this (also by Euler's method)" is meaningless. You need to proceed as I indicated in #15.

    Now, what you wrote after "Also, someone suggested this (also by Euler's method)" makes a lot more sense. These are (almost) the equations you should obtain by following my earlier advice. Two things are wrong: the signs in the Ax and Ay terms, and the magic number 39.8 - where is it even coming from?
     
  19. Aug 19, 2014 #18
    I think 39.8 is initial centripetal acceleration + gravitational acceleration.
    The equation for the magnitude of A_L (meaning acceleration due to lift) is A_L = (30+g)(V_1/60)^2. Note that it includes g - this is because the amount of lift at the start must provide enough force to overcome gravity plus accelerate upwards at 30 m/s^2.

    Does the following make sense? :

    θ = ω * t ... angle between the position of the plane in the vertical going through the center of the loop

    2yv1h5v.jpg
     
  20. Aug 19, 2014 #19
    OK, that seems correct. But the signs are still wrong. The force of drag is always against the velocity.

    That above does make sense to me, even though I think it is wrong. The real question here is, does that make sense to you?
     
  21. Aug 19, 2014 #20
    I try to understand but it's very hard for me. This above are equations of motion under the constrain of the centripetal force. But unfortunately, I think this is not going with Euler's method. My assignment is to program in C# with Euler's method. I think programming won't be so difficult when I get this physics theory (equations) right. But this is such a struggle :(
     
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