Modelling The Landing Of An Airplane

Before creating a model for the airplane's landing, I decided to make a velocity-time graph; using the data I was given.

The graph shows that there is a significant change in the deceleration of the airplane after t = 9. This suggests that the journey of the aircraft is seperated into two distinct phases, both of which require separate models. Practically speaking, this is when the breaks would likely have been applied, since the gradient suddenly steepens, which shows us that the deceleration increased, therefore the extra (breaking) force must have been applied at this point to slow down the airplane. This is in addition to the existing air resistance which was initially the only force assumed to be acting on it, assuming that ground friction at the initial stage is negligible.

Initially, I will form a linear drag model for the airplane landing such that:

FD = kv

Where FD is the drag force, v is the velocity of the airplane, and k is a constant to be determined.

This leads to the following differential equation to be solved for v :

Where A is another constant to be determined.

Substitution initial conditions gives us the following:

t = 0, v = 96

t = 9, v = 55, m = 120,000 (from data)

, 0 ? t ? 9

The following graph demonstrates the closeness of fit this model, with the actual data provided to us on the airplane's velocities in this period of time.

Whilst the fit isn't too far off, there is still a significant gap between the data values, and so a more sophisticated model is necessary, since this linear drag model is only accurate at low velocities. To model the air resistance solely acting upon the airplance from t=0 to t=9, we can now use the revised drag equation to give an improved model of the landing trajectory, which is as follows:

FD = ½ ?v2CDA

Where FD is the drag force in direction of flow velocity, ? is the mass density of the fluid (in this case air), v is the velocity of the object (airplane) relative to the fluid, CD is the drag co-efficient which depends on the shape of the moving object, and finally A is the reference area (the area of the object which the drag force is applied to). Since the density of the air is constant at ground level, and the shape of the airplane doesn't change (the flaps are already extended prior to landing), our former variables ?, CD and A can now be regarded as all constant. The velocity v is therefore the only factor which varies, and so this now simplifies our drag equation to:

FD = kv2

We can now form a differential equation to be solved for v which is as follows:

Before continuing, I shall explain the reasoning for above. There is a minus sign in front of the k because the drag is opposing the motion of the aircraft, and m dv/dt represents the force of the aircraft's motion, with m being the aircraft of mass 120,000 kg, and dv/dt representing its deceleration. To determine our constants k and c, we must now substitute two initial conditions to determine what they are, which now follows.

From our data, when t=0, v=96, so we shall substitute these into the formula, this gives:

Next I will substitute t=9, v=55, m= 120,000 and c = -1/96 which yields:

We now have our model for the first part of the flight's landing, from t=0 to t=9:

, k = 10250/99, m = 120,000, c = -1/96

, 0 ? t ? 9

The model fitted the calculated data extremely well, with the theoretical values being exactly the same as the actual data when rounded to 2 significant figures. This is clearly shown by the graph below:

Although the model fitted the data very well in the range t= 0 to t=9 (the trend lines overlapped almost perfectly), it is limited by the fact that had the airplane only been subjected to air resistance throughout its entire landing, as it's velocity will never reach 0 due to the asymptotic nature of this model. Conveniently we have our next section of the model which one would hope to result in a real corresponding t value for v = 0.

The next section of the plane's landing, 9 ? t ? 26, now involves an extra frictional force since the airplane's brakes are now applied. We must assume this as constant, since we are not provided with any other modelling information. This now adds in addition to the pre-existing air resistance, yielding the following differential equation:

This is with k being the same constant as before (10250/99), since the air-resistance hasn't changed, and B now represents the additional frictional force. The capital letter T represents the time T seconds after the brakes were applied. This can then be solved as follows for 0 ? T ? 19:

Substituting the initial conditions v = 0, when T=17 yields:

(1)

The above is now our first simultaneous equation for d, which is another arbitrary integration constant.

Next substituting v = 55 when T = 0 gives us:

(2)

(1) x (2) =

, k = 10250/99, m=120000

The above equation for d can now only be solved using a numerical method.

I will use the Newton-Raphson method by considering:

I drew a graph of the function so that I could gain a rough idea on the location of the principal value of its solution, in order for me to pick an appropriate starting value x0 which lies near the root, so that my iteration converges to it. From it I discovered that the root lied between 0.5 and 1, so I picked my initial value as 0.7.

I then input it into the following iteration formula:

, substituting my values of f(x) and f'(x) obtained above gave me

The following table demonstrates the iterations which were made:

n xn 0 0.7 1 0.80648426 2 0.793825758 3 0.793595153 4 0.793595079 5 0.793595079 6 0.793595079 7 0.793595079 8 0.793595079 9 0.793595079 10 0.793595079

My value of d converged to d = 0.793595079. Substituting this back into equation (1) also gives us:

, which is our frictional force exerted by the brakes.

I can now substitute my values into my equation for v to gain a particular solution to the differential equation which now follows:

,

B = 303,091.6073, m = 120,000, k = 10,250/99, d = 0.793595079.

I will also now substitute T with T = t-9, since I want this function to continue on from where the previous function ended at t = 9, so that it models the airplane's behaviour in the section 9 ? t ? 26 of its journey.

, 9 ? t ? 26

Once again, the model fitted the calculated data very closely, with the theoretical values being exactly the same as the actual data when rounded to the nearest integer value. The graph below demonstrates this, as once again the lines of the theoretical and actual velocities almost overlap perfectly. Furthermore this model is sufficient in describing the remainder of the airplane's journey because its velocity eventually reaches v = 0 as it does in real life, with no asymptote lying here.

Finally, I am now going to use the trapezium rule on the data I was originally given, so that I can recommend a minimum length for this runway, since finding the area under a velocity time graph gives us the airplane's displacement S upon landing. To do this I must also assume that this displacement represents motion in a straight line. I must also assume that this runway is only being used for landing and not taking off, as this would require a significantly greater length in order to allow the airplane to reach sufficient speed for take-off.

The smallest intervals my data has is 1, so I shall make h = 1 to use in the trapezium rule. I will then apply it to the data as follows:

S = ½ x (96 + 0 + 2 x (89 + 82 + 77 + 72 + 68 + 64 + 61 + 58 + 55 + 50 + 46 + 41 + 38 + 34 + 31 + 27 + 24 + 21 + 18 + 16 + 13 + 10 + 8 + 5 + 3)) = 1059 metres.

This is an absolute bare minimum for the runway length, which assumes that the brakes will work fully. In the worst case scenario where the brakes were to not work at all after t = 9, I will extend the first model to an arbitrary time t, which gives a value for v sufficiently close to 0 when I can assume the airplane is no longer moving. At t = 250, the airplane is moving less than 5 ms-1, and so factors such as ground friction which were previously neglected would now be significant enough to slow the airplane down completely. Integrating the first function between t = 0 and t = 250 gives me a runway length as follows:

= 3567 m (to the nearest metre)

The above estimate should be more than safe enough for a runway length, as it enables sufficient time to pass for the airplane to slow down to facilitate a rescue, in case the brakes stop working unexpectedly.