# Homework Help: Modelling an Aeroplane - Non-Linear Differential Equation

1. Apr 8, 2010

### MaceLee

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:
$$m\ddot{x} = -k\dot{x}^2$$

I'm trying to find $$\ddot{x}$$, but I'll probably need to find $$x$$ as well

3. The attempt at a solution

$$v = \dot{x}$$

$$m\dot{v} = -kv^2$$

$$\dot{v} = -(k/m)v^2$$

$$-(\frac{m}{k})\int\frac{1}{v^2} dv = \int dt$$

$$\frac{m}{kv} + c = t$$

$$\dot{x} = v = \frac{m}{k(t-c)}$$

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. Apr 8, 2010

### espen180

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.

3. Apr 8, 2010

### MaceLee

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!

4. Apr 8, 2010

### RoyalCat

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:

$$m\ddot x = -k \dot x ^2$$

$$\dot v = -\gamma v^2$$, $$\gamma \equiv \frac{k}{m}$$

$$\frac{dv}{dt}=-\gamma v^2$$

$$\frac{dv}{v^2}=-\gamma dt$$

$$\int^v_{v_0} \frac{dv}{v^2}=-\gamma \int^t_0 dt$$

$$\frac{1}{v_0}-\frac{1}{v} = -\gamma t$$

$$v=\frac{v_0}{1+v_0\gamma t}$$

Your answer is correct provided that $$c=-\frac{1}{\gamma v_0}$$

5. Apr 8, 2010

### MaceLee

I see; your solution is indeed much neater. Very much appreciated, I'll remember that in future :) Thanks

6. Apr 9, 2010

### MaceLee

I've just moved onto the second equation and I'm trying to solve it:

$$\dot{v} = -3\lambda v^2 - \mu$$

Where $$\lambda$$ and $$\mu$$ are known constants.

I tried to solve it as follows:

$$3\lambda \int^v_{v_0}v^2 dv = -\mu\int^t_{0}dt$$

Giving:
$$\lambda v^3 - \lambda v_0^3 = -\mu t$$

Resulting in:
$$v^3 = -\frac{\mu}{\lambda} t + v_0^3$$

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

7. Apr 10, 2010

### MaceLee

I've been trying for the past day but still can't solve this equation:

$$\dot{v} = -\gamma v^2 - \lambda$$

Where $$\gamma$$ and $$\lambda$$ are known constants

Thanks!

Last edited: Apr 10, 2010
8. Apr 10, 2010

### RoyalCat

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 $$\frac{d(\tan{x})}{dx} = 1+\tan^2{x}$$ and then it should all come together. Guessing a solution of the form $$v=k\tan{(at + b)}$$ 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 $$\mu$$ ?
Replacing it with a plus makes the solution make sense, it turns the $$\tan$$ into a $$\tanh$$ 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:

$$m\dot v = -kv^2 + g$$

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

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 $$\sqrt{\mu}$$ you have $$\sqrt{-\mu}=i\sqrt{\mu}$$ which turns the tan into tanh, so it is just a matter of a constant. ;))

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

Which approaches a terminal velocity $$v_t=\sqrt\frac{\mu}{\gamma}$$ as expected.

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

Best of luck with your studies. :)

Last edited: Apr 10, 2010
9. Apr 11, 2010

### ehild

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

So this is the ODE:

$$\dot{v} = -3\lambda v^2 - \mu$$

It is a first order and separable one:

$$\frac{dv}{dt} = -\mu(\frac{3\lambda}{\mu} v^2 + 1)$$

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

With the new variable

$$x=\sqrt{\frac{3\lambda}{\mu}} v$$

the integral is the form of

$$\int{\frac{1}{x^2+1}dx}= \arctan {x}$$

with the notation

$$k=\sqrt{\frac{3\lambda}{\mu}}$$

the solution is:

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

taking the tangent of both sides:

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

ehild

Last edited: Apr 11, 2010
10. Apr 11, 2010

### MaceLee

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.

:)

11. Apr 11, 2010

### MaceLee

Just one last thing! Taking the tangent of both sides gives:

$$kv(t) = kv_0 - tan(\mu t)$$

Taking the k to the other side gives:

$$v(t) = v_0 - \frac{1}{k} tan(\mu t)$$

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

Thanks!

12. Apr 11, 2010

### ehild

No.

$$\tan{(\alpha-\beta)}=\frac{\tan{\alpha}-\tan{\beta}}{1+\tan{\alpha}\tan{\beta}}$$

ehild

13. Apr 11, 2010

### MaceLee

Ah silly mistake, I took the tan separately. Thanks again ehild.

14. Apr 11, 2010

### ehild

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

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

so the solution is:

$$\arctan{kv(t)} = \arctan{k v_0}-\mu k t$$

taking the tangent of both sides:

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

I hope this is correct.

ehild

Last edited: Apr 12, 2010
15. Apr 12, 2010

### MaceLee

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 $$v_0$$, 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?

16. Apr 12, 2010

### ehild

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

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

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

ehild

Last edited: Apr 12, 2010
17. Apr 12, 2010

### MaceLee

$$\mu = \frac{301687.25}{120000}$$

$$3 \lambda = 103.02$$

$$v_0$$ was set as to satisfy when t = 10, v = 50

18. Apr 12, 2010

### ehild

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:

• ###### aeroplane.jpg
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19. Apr 12, 2010

### MaceLee

Oh yes, $$3 \lambda$$ was written incorrectly, it should have been:

$$3 \lambda = \frac{103.02}{120000}$$

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

20. Apr 12, 2010

### ehild

With this 3lambda, everything fits!!!

ehild

21. Apr 12, 2010

### MaceLee

Just tried it all in excel. It gives great results :D Thank you very much ehild! Life saver :)

22. Apr 12, 2010

### RoyalCat

I think I found the source of your mistake. You used the velocity when the braking began to extrapolate the velocity at t=0, and you called that velocity $$v_0$$ This is incorrect.
You assumed that the plane had been breaking throughout t=0 to t=10, but that might not be the case. Either way, you should take $$v_0$$ as the velocity the moment the plane started breaking, and then use $$t'=t-t_0$$ for your new time parameter, since what you care about is the process once the plane has started breaking.

23. Apr 12, 2010

### ehild

You are right, the breaking began at t=10, so I took it into account in my fitting function (used t=time-10 instead of the given time values in the fitting function), and it worked surprisingly well! But the fitting works with the original t values, too, only vo is much higher (147) then.

ehild

Last edited: Apr 12, 2010
24. Apr 13, 2010

### MaceLee

Ah, indeed. Prior to being told about using limits, I had no idea that you could do so in separation of variables - so I assumed it would work the same way in all cases. Regardless, the overall result an equation which fits the data well; $$v_0$$ ends up as you said, around 140.

I've just printed all my work; would you say it's worth making the changes and reprinting? Would I be marked down for solving for $$v_0$$ (bearing in mind this is A-level)?

Many Thanks :)

25. Apr 13, 2010

### ehild

The v(t) function is in terms of the "braking time" which is the indicated time-10 s. So Vo is really V10. You can either plot the data in terms of the braking time and then vo = 50 m/s, or keep the original time values, but calculate the velocities in term of the braking time. If so, denote vo by v(10). That 147 as vo has no sense, as the velocity function was different before the braking started.

ehild