1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Modelling an Aeroplane - Non-Linear Differential Equation

  1. Apr 8, 2010 #1
    Hi everyone,

    1. The problem statement, all variables and given/known data

    I've been having a bit of trouble with my mathematics coursework. We have to model an aeroplane that has just landed with two equations. One prior to braking, one after. I've used a model with a quadratic drag approximation; and I'm not certain if my result is correct. Mainly because we haven't covered anything to do with non-linear differential equations, and suddenly I've found myself with one.


    2. Relevant equations

    I'm modelling the system before a constant braking force with this DE:
    [tex]m\ddot{x} = -k\dot{x}^2[/tex]

    I'm trying to find [tex]\ddot{x}[/tex], but I'll probably need to find [tex]x[/tex] as well

    3. The attempt at a solution

    [tex]v = \dot{x}[/tex]

    [tex]m\dot{v} = -kv^2[/tex]

    [tex]\dot{v} = -(k/m)v^2[/tex]

    [tex]-(\frac{m}{k})\int\frac{1}{v^2} dv = \int dt[/tex]

    [tex]\frac{m}{kv} + c = t[/tex]

    [tex]\dot{x} = v = \frac{m}{k(t-c)}[/tex]

    Is this correct? All the constants I'll find by trial and error matching it up to the velocity-time data we were given.

    Thanks :)
     
    Last edited: Apr 8, 2010
  2. jcsd
  3. Apr 8, 2010 #2
    If you are given a table of time and velocity data, you should plot these against each other. You should be able to see from the plot whether your proposed equation is plausible.
     
  4. Apr 8, 2010 #3
    I just tried that out, it would seem that the equation isn't plausible. I'll give it another go with linear drag instead.

    Thanks!
     
  5. Apr 8, 2010 #4
    When you solve a differential equation using integration, you should always take note of the limits of integration. Your solution will get very messy one you start trying to get c out, as opposed to using the limits of integration and getting it directly:

    [tex]m\ddot x = -k \dot x ^2[/tex]

    [tex]\dot v = -\gamma v^2[/tex], [tex]\gamma \equiv \frac{k}{m}[/tex]

    [tex]\frac{dv}{dt}=-\gamma v^2[/tex]

    [tex]\frac{dv}{v^2}=-\gamma dt[/tex]

    [tex]\int^v_{v_0} \frac{dv}{v^2}=-\gamma \int^t_0 dt[/tex]

    [tex]\frac{1}{v_0}-\frac{1}{v} = -\gamma t[/tex]

    [tex]v=\frac{v_0}{1+v_0\gamma t}[/tex]

    Your answer is correct provided that [tex]c=-\frac{1}{\gamma v_0}[/tex]
     
  6. Apr 8, 2010 #5
    I see; your solution is indeed much neater. Very much appreciated, I'll remember that in future :) Thanks
     
  7. Apr 9, 2010 #6
    I've just moved onto the second equation and I'm trying to solve it:

    [tex]\dot{v} = -3\lambda v^2 - \mu[/tex]

    Where [tex]\lambda[/tex] and [tex]\mu[/tex] are known constants.

    I tried to solve it as follows:

    [tex]3\lambda \int^v_{v_0}v^2 dv = -\mu\int^t_{0}dt[/tex]

    Giving:
    [tex]\lambda v^3 - \lambda v_0^3 = -\mu t[/tex]

    Resulting in:
    [tex]v^3 = -\frac{\mu}{\lambda} t + v_0^3[/tex]

    But looking at my data, this appears to be incorrect. Plotting v^3 against t does not result in a straight line. I gave wolfram alpha a go, but it gave me a weird solution involving tan. Have I done something incorrectly?

    Thanks
     
  8. Apr 10, 2010 #7
    I've been trying for the past day but still can't solve this equation:

    [tex]\dot{v} = -\gamma v^2 - \lambda[/tex]

    Where [tex]\gamma[/tex] and [tex]\lambda[/tex] are known constants

    Could anyone help me please?

    Thanks!
     
    Last edited: Apr 10, 2010
  9. Apr 10, 2010 #8
    The jump from the first line to the second has numerous mistakes, review it and find your mistakes.

    I've found the same solution using Wolfram as well, but I'm afraid the ODE is a bit out of my range.
    Now that we know that it goes like tan, then we can rationalize it. One way of 'guessing' the form of the solution, is noting that [tex]\frac{d(\tan{x})}{dx} = 1+\tan^2{x}[/tex] and then it should all come together. Guessing a solution of the form [tex]v=k\tan{(at + b)}[/tex] is a logical step to take.

    As far as my limited knowledge goes, for these kinds of ODE's (Non-linear, inhomogeneous), there's not much you can do but throw everything you have at them, integrals, guesses at the solution, what have you. There's little method to the madness. :)

    Though someone more clever than I am may know of a way of solving this particular kind of ODE off the top of his head. ;) I'd love to see that solution.

    On a side note, are you sure that you need a minus in front of the [tex]\mu[/tex] ?
    Replacing it with a plus makes the solution make sense, it turns the [tex]\tan[/tex] into a [tex]\tanh[/tex] which has the asymptotic approach to a terminal velocity which we ought to expect.

    I'm assuming that the equation you've been trying to solve is freefall under the effect of gravity and a retarding force proportional to the square of the velocity.

    Taking the positive direction as towards the ground, the equation of motion reads:

    [tex]m\dot v = -kv^2 + g[/tex]

    [tex]\dot v = -\gamma v^2 + \mu[/tex] where we have defined [tex]\frac{k}{m} \equiv \gamma, \mu \equiv \frac{g}{m}[/tex]

    A mistake of sign in a differential equation stands for much more than just a different constant in the final solution.. (Though here you can see that instead of [tex]\sqrt{\mu}[/tex] you have [tex]\sqrt{-\mu}=i\sqrt{\mu}[/tex] which turns the tan into tanh, so it is just a matter of a constant. ;))

    Wolfram's answer for the above ODE is:
    [tex]v[t]\to \frac{\sqrt{\mu } \text{Tanh}\left[t \sqrt{\gamma } \sqrt{\mu }+\text{ArcTanh}\left[\frac{u \sqrt{\gamma }}{\sqrt{\mu }}\right]\right]}{\sqrt{\gamma }}[/tex]

    Which approaches a terminal velocity [tex]v_t=\sqrt\frac{\mu}{\gamma}[/tex] as expected.

    And this too is something we could have guessed, noting that [tex]\frac{d(\tanh{x})}{dx}=1-\tanh^2{x}[/tex]

    Best of luck with your studies. :)
     
    Last edited: Apr 10, 2010
  10. Apr 11, 2010 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    It is the braking after landing, as written in the first post.

    So this is the ODE:

    [tex]
    \dot{v} = -3\lambda v^2 - \mu
    [/tex]

    It is a first order and separable one:

    [tex]
    \frac{dv}{dt} = -\mu(\frac{3\lambda}{\mu} v^2 + 1)
    [/tex]

    [tex]
    \int_{v_0}^{v(t)} \frac{dv}{(\frac{3\lambda}{\mu} v^2 + 1)} = -\mu t
    [/tex]

    With the new variable

    [tex]x=\sqrt{\frac{3\lambda}{\mu}} v [/tex]

    the integral is the form of

    [tex] \int{\frac{1}{x^2+1}dx}= \arctan {x} [/tex]

    with the notation

    [tex]k=\sqrt{\frac{3\lambda}{\mu}}[/tex]

    the solution is:

    [tex] \arctan{kv(t)} = \arctan{kv_0}-\mu t[/tex]

    taking the tangent of both sides:

    [tex]v(t)=\frac{v_0-1/k \tan{\mu t}}{1+kv_0 \tan{\mu t}} [/tex]

    ehild
     
    Last edited: Apr 11, 2010
  11. Apr 11, 2010 #10
    Thank you so much, ehild & RoyalCat. I had tried by differentiating it and getting a solution from the second order linear - it works for a few values but then suddenly goes wildly astray.

    :)
     
  12. Apr 11, 2010 #11
    Just one last thing! Taking the tangent of both sides gives:

    [tex]kv(t) = kv_0 - tan(\mu t)[/tex]

    Taking the k to the other side gives:

    [tex]v(t) = v_0 - \frac{1}{k} tan(\mu t)[/tex]

    Where did the denominator come from in your solution? Was it a mistake? Or did I do something wrong again :/ ?

    Thanks!
     
  13. Apr 11, 2010 #12

    ehild

    User Avatar
    Homework Helper
    Gold Member


    No.

    [tex]\tan{(\alpha-\beta)}=\frac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}}[/tex]

    ehild
     
  14. Apr 11, 2010 #13
    Ah silly mistake, I took the tan separately. Thanks again ehild.
     
  15. Apr 11, 2010 #14

    ehild

    User Avatar
    Homework Helper
    Gold Member

    I made a mistake. After introducing the new variable x=kv, dv=dx/k, we have the equation

    [tex]


    \int{\frac{dx}{x^2+1}} = - \mu k t


    [/tex]

    so the solution is:

    [tex] \arctan{kv(t)} = \arctan{k v_0}-\mu k t [/tex]

    taking the tangent of both sides:

    [tex]v(t)=\frac{v_0 - 1/k \tan{\mu k t}}{1+kv_0 \tan{ \mu k t} [/tex]

    I hope this is correct.


    ehild
     
    Last edited: Apr 12, 2010
  16. Apr 12, 2010 #15
    Thanks ehild. I tried to use the solution to find predictions for experimental data, but the values are extremely strange. I'm not sure if it's the way I've worked out [tex]v_0[/tex], but seemingly regardless of the value, the output jumps around really weirdly. The data I'm supposed to be modelling with the equation is:

    t v
    10 50
    11 46
    12 41
    13 38
    14 34
    15 31
    16 27
    17 24
    18 21
    19 18
    20 16
    21 13
    22 10
    23 8
    24 5
    25 3
    26 0

    But the actual values that come out as v are more on the lines of:

    49.93375881
    -11.79891637
    -129.3575502
    85.61016673
    2.555699999
    -70.00352361
    173.8124513
    17.27837342
    -41.21109751
    1353.425599
    34.75669316
    -22.1799659
    -239.8506759
    59.50383843
    -6.916230932
    -102.9952062
    104.9325927

    It jumps from negative to positive pretty frequently. Would this be caused by working the constant term out incorrectly?
     
  17. Apr 12, 2010 #16

    ehild

    User Avatar
    Homework Helper
    Gold Member

    What values did you use for mu, lambda, vo?

    It would be nice of you checking my solution. I am unlucky with that k, putting it at the wrong place. So my new version is
    [tex]


    \int{\frac{kdx}{x^2 + 1} = -k \mu t


    [/tex]

    [tex]

    \arctan{kv(t)} = \arctan{kv_0}-\mu k t

    [/tex]


    [tex]

    v(t)=\frac{v_0 -1/k \tan{\mu k t }}{1+kv_0 \tan{\mu k t}}

    [/tex]

    ehild
     
    Last edited: Apr 12, 2010
  18. Apr 12, 2010 #17
    [tex]\mu = \frac{301687.25}{120000}[/tex]

    [tex]3 \lambda = 103.02[/tex]

    [tex]v_0[/tex] was set as to satisfy when t = 10, v = 50
     
  19. Apr 12, 2010 #18

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Where did these value come from? mu is all right, but 3lambda seems too big. You know that the tangent becomes infinite at pi/2, so k*mu*t cannot be greater than that. And the solution is valid only for positive
    velocities. If the tangent is too big, v becomes negative. That constant force is realistic only till the aeroplane stops.

    In what units did you measure the speed and the time? Did the braking start at t=10?
    I used your data and the last version of my function to get the parameters by fitting. See the attached picture. But I got vo (actually it is at t=10) 50.2, mu =2.504, 3lambda=8.86˙10-4.

    ehild
     

    Attached Files:

  20. Apr 12, 2010 #19
    Oh yes, [tex]3 \lambda[/tex] was written incorrectly, it should have been:

    [tex]3 \lambda = \frac{103.02}{120000}[/tex]

    Sorry about that D: The units are standard, v in ms^-1 and t in seconds. The breaking started at t=10 onwards yes.

    Your attached graph is a really close fit! I'll try out your new function now with these values :)

    Thanks
     
  21. Apr 12, 2010 #20

    ehild

    User Avatar
    Homework Helper
    Gold Member

    With this 3lambda, everything fits!!!

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Modelling an Aeroplane - Non-Linear Differential Equation
Loading...