Dierential Equations Coursework - Aeroplane Landing

piguy.org

March 14th, 2014

Contents

1 Introduction 2

2 Simplifying and Setting up the Model 22.1 Assumptions Made . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Relevance of Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.3 Establishing Dierential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 Manipulating the Model 43.1 Solving the Dierential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43.2 Choosing the parameters of the DE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43.3 Particular Solutions to the Dierential Equations . . . . . . . . . . . . . . . . . . . . . . . 53.4 Predictions Produced from Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

4 Verifying the Model 64.1 Collecting the data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64.2 The Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

5 Comparison of Predictions and Data 75.1 Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75.2 Variations of Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

6 Revision of the Model 86.1 Ammendments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86.2 Establishing New Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

7 Assesment and Conclusions 97.1 Solving the New Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97.2 Conclusion Comparing New Predictions and Data . . . . . . . . . . . . . . . . . . . . . . 107.3 Recommending minimum runway length . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1

EXAM

PLE

1 Introduction

The task, Aeroplane Landing, requires the modelling of the velocity of a landing aeroplane from thetime it touches down to coming to a rest in the following situation:

An aeroplane of mass 120,000kg comes in to land at a runway. After touchdown it initiallyslows down from air resistance. When it has slowed down enough this is augmented by aconstant force from the wheel brakes.

We are to investigate a suitable mathematical model using Dierential Equations to explain the natureof the forces acting on the aeroplane during these 26 seconds.

The task has already provided us with data for the velocity each second from 0 to 26 to aid with themodelling cycle, which will be discussed later in order to test our models. (see Verifying the Model)

2 Simplifying and Setting up the Model

2.1 Assumptions Made

There are a number of assumptions made in the model, listed in descending order of relative importance:

1. The aerodynamics of the plane do not change. That is, the plane does not deploy any flaps,ailerons or elevators during landing. This is the most signficant since it has a direct consequencefor the air resistance experienced by the plane.

2. The aeroplane acts as a point. This simplifies the complex aerodynamics and drag by suggestingthat we can consider the plane as a single point at its centre of mass rather than as a whole body.

3. There is no wind speed on the ground. We dont know if there is any wind going againstthe plane - which would act against motion and contribute to its deceleration, or with the plane,which would work against deceleration. However we dont know what the conditions are, so it isbest to assume there is no headwind - which in reality is unlikely.

4. The runway is completely level. Even with very small slopes, the component of the 120,000gweight would still be very large in magnitude to be able to aect the deceleration largely.

5. The landing is smooth. If the landing was rough, it would cause the normal reaction force tovary as the plane landed, which would aect friction and hence the constant braking force.

6. The surface of the runway is uniform. Since were considering a constant braking force, itdoesnt matter exactly what the surface is, so long as it doesnt change - in reality there could bedebris or oil on sections which aect the coecient, though it would be minor.

7. The landing is completely straight. If the plane were to be turning while landing, it wouldincrease the friction of the tires and we would need to consider the force needed to change thedirection component of its velocity.

8. The fuel burnt during landing is negligible. This would mean that the mass of the planedecreased, hence its inertia and resistance to deceleration decreased. However in 26 seconds, thefuel burnt will be a tiny proportion of its 120,000kg mass.

9. The weather conditions do not change. This would eect the surface of the runway, windspeeds (see assumptions 2 & 5) as well as possibily the density of air, which would aect airresistance, however any changes to the weather within 26 seconds would be very minimal.

10. Relativistic eects are negligible. Even at the maximum velocity of 96m/s, it is very small incomparison to the speed of light, 3 108m/s, so changes to mass and the eect of time dilationwould be incredibly small, infact it is likley immeasurable.

2

EXAM

PLE

2.2 Relevance of Assumptions

From our assumptions #2, #3 and #4, we can establish a model of the forces:

The first case is before the braking force is applied; the second model adds this force. Since assumptions1 & 2 simplify the aerodynamics of the plane to be considered a single point, we note that the dragforce due to air resistance, Fdrag / v, where v is the velocity of the aeroplane. From the rest of ourassumptions, we can just take a constant of proportionality without worrying about additional constantsneeding to be added, so Fdrag = kv.

Also since the mass is not changing from assumptions 7 and 10, we can also use the special case ofNewtons Second Law, that F = ma, rather than F = d(mv)dt .

2.3 Establishing Dierential Equations

We consider the model in two cases; without and with the braking force.

Case 1 Plane initially lands, with only air resistance providing deceleration force.

Fnet = ma

=) Fdrag = mdvdt

=) mdvdt

= kv (1)

Case 2 Braking force is applied.

=) Fdrag Fbraking = mdvdt

=) mdvdt

= kv FB (2)

3

EXAM

PLE

3 Manipulating the Model

3.1 Solving the Dierential Equations

We can solve this first order dierential equation using the seperation of varibles:

mdv

dt= kv

=)Z

1

vdv =

Z km

dt

=) ln|v| = kmt+ c (3)

We can use integrating factors to solve our second dierential equation:

mdv

dt= kv FB

dv

dt+

k

mv = FB

Multiplying both sides of the equation by the integrating factor, R = eRP (t) dt, where P (t) = km

ekmdv

dt+ e

ktm

k

mv = e ktm FB

=) ddt

ektm v

= e ktm FB

=) e ktm v =Ze ktm FB dt

=) e ktm v = kmFB e ktm + c

=) v = ce km t kmFB (4)

3.2 Choosing the parameters of the DE

We need to identify the initial conditions of the dierential equations to get particular solutions. Fromthe data given (see section 4), we know the motion is split into before and after the brakes are applied.

From section 4.2, we decided that this occured at t = 10, so our initial conditions are:

case 1 - t = 0, v = 96m/s; t = 9, v = 55m/scase 2 - t = 10, v = 50m/s; t = 26, v = 0m/s

However these intial conditions can vary, since the data we are given is to the closest integer value, thusthere is an error of 0.5 for all points used. For instance in case one, the upper bound t = 0, v = 96.5m/sand t = 8.5, v = 55.5m/s, giving a percentage error of (0.5/9 + 0.5/55) 100% = 6.4%

4

EXAM

PLE

3.3 Particular Solutions to the Dierential Equations

For equation (3), we know that at t = 0, v = 96m/s; also at t = 9, v = 55m/s. Thus we obtainsimulaneous equations:

ln|96| = c , ln|55| = km9 + c

Hence we obtain the particular solution that c = ln96 and that km = ln(5596 )/9, giving:

v = exp (ln96 0.06189t) (5)

For equation (4) modelling the velocity after the brakes have been applied, we now know that km =ln( 5596 )/9 as well as our parameters that when t = 26, v = 0m/s; also at t = 10, v = 50m/s. Thus weobtain simulaneous equations:

ln

55

96

FB9

= c.exp

ln

55

96

26

9

ln

55

96

FB9

= c.exp

ln

55

96

10

9

50

Hence we obtain the particular solution that c = 15.915 and that FB = 1285.4, giving:

v = 79.55 15.915e0.06189t (6)

3.4 Predictions Produced from Equations

t (s) 0 1 2 3 4 5 6 7 8 9v (m/s) 96 90 85 80 75 70 66 62 59 55

t (s) 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26v (m/s) 50 48 46 44 42 39 37 34 31 28 25 21 17 14 9 5 0

0 2 4 6 8 10 12 14 16 18 20 22 24 260

20

40

60

80

100

time (s)

speed(m

/s)

5

EXAM

PLE

4 Verifying the Model

4.1 Collecting the data

We have already been supplied with the times and velocities of the aeroplane with the task. As discussedin 3.2, there is quite a large amount of uncertainty in the data, since we are given it to the closest integervalue, thus there is an error of 0.5 for all points used. For instance in case one, the upper bound t = 0,v = 96.5m/s and t = 8.5, v = 55.5m/s, giving a percentage error of (0.5/9 + 0.5/55) 100% = 6.4%

4.2 The Data

The following table of results gives the speed v, t seconds after touch down:

0 2 4 6 8 10 12 14 16 18 20 22 24 260

10

20

30

40

50

60

70

80

90

100

time (s)

speed(m

/s)

t (s) 0 1 2 3 4 5 6 7 8 9v (m/s) 96 89 82 77 72 68 64 61 58 55

t (s) 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26v (m/s) 50 48 46 44 42 39 37 34 31 28 25 21 17 14 9 5 0

Clearly we observe a dierence in the gradient of the curve between t = 9 and t = 10 seconds, whichindicates that this was the point at which the braking force was applied since the rate of decelerationincreases after t = 10. Since we do not know specifically at which point between 9 and 10 the planestarted to brake, we can only assume that at t = 10 is the inital point at which the model for brakingforce begins.

6

EXAM

PLE

5 Comparison of Predictions and Data

5.1 Comparison

0 2 4 6 8 10 12 14 16 18 20 22 24 260

10

20

30

40

50

60

70

80

90

100

time (s)

speed(m

/s)

t (s) 0 1 2 3 4 5 6 7 8 9vactual (m/s) 96 89 82 77 72 68 64 61 58 55vpredicted (m/s) 96 90 85 80 75 70 66 62 59 55

t (s) 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26vactual (m/s) 50 46 41 38 34 31 27 24 21 18 16 13 10 8 5 3 0vpredicted (m/s) 50 48 46 44 42 39 37 34 31 28 25 21 17 14 9 5 0

5.2 Variations of Parameters

Our initial model clearly doesnt fit very well at all with the recorded data we were given, particularly inthe latter part of the motion when the braking force was also applied, although the earlier part for thefirst 9 seconds fits within the standard error bounds of the data we were given, since our initial parameterscould vary 6.4%. However later predicted data falls well out of range with the actual velocities, seen inthe table:

t (s) 0 1 2 3 4 5 6 7 8 9change in v 0 1 3 3 3 2 2 2 1 0max change 6 6 5 5 5 5 4 4 4 4

7

EXAM

PLE

t (s) 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26change in v 0 2 5 6 8 7 10 10 10 10 9 8 7 6 4 2 0max change 4 3 3 2 2 2 2 1 1 1 1 1 1 0 0 0 0

6 Revision of the Model

6.1 Ammendments

We could improve on the model, which has been seen to be not quite perfect, by relaxing one of oursignificant simplyfying assumptions given in 2.1.

We can relax assumption #2, that the plane acts as a single point in space. If we consider it to occupyspace in three dimensions, this alters our original drag equation, as now air resistance acts at a cross-sectional area, A, of the plane, rather than a single point.

So now in time, t, the plane covers a distance:

s = vt

Hence the volume of air, V , that it covers is

V = Avt

Since = mV , this means that:m = V = Avt

Also since the air is at rest (assumption #3) but is then accelerated to the velocity of the plane, v whenit hits the plane, then by Newtons second law:

aair =v

t=

v

t

=) Fdrag = ma = Avtvt

= Av2 = kv2

So we will ammend our new model to suppose that Fdrag = kv2, rather than our ealier assumption thatthere was a linear relationship: Fdrag = kv.

6.2 Establishing New Equations

We consider the model in two cases; without and with the braking force.

Case 1 Plane initially lands, with only air resistance providing deceleration force.

Fnet = ma

=) Fdrag = mdvdt

=) mdvdt

= kv2 (7)

Case 2 Braking force is applied.

Fdrag Fbraking = mdvdt

=) mdvdt

= kv2 FB (8)

8

EXAM

PLE

7 Assesment and Conclusions

7.1 Solving the New Equations

We can solve equation (7) using normal separation of variables:

mdv

dt= kv2

=)Z

1

v2dv =

Z km

dt

=) 1v= k

mt+ c

=) 1v=

k

mt+ c (9)

we know that at t = 0, v = 96m/s; also at t = 9, v = 55m/s. After solving the simultaneous equationswe get that k/m = 8.628 104, and the particular solution:

1

v= 8.628 104t+ 0.01042 (10)

We now need to solve the other dierential equation:

mdv

dt= kv2 FB

=)Z 1

kv2 + FBdv =

Z1

mdt

=) 1k

Z1

v2 + FBkdv =

Z1

mdt

=) 1darctan

vd

=

kt

m+ c

where d =q

FBk

We know that at t = 26, v = 0m/s, thus we see that c = 0.0224328.To solve the remaining equation for FB , we will make use of the Newton-Raphson iteration for theremaining simultaneous equation which occurs when t = 10, v = 50m/s:

=) 1darctan

50

d

0.0138048 = 0

9

EXAM

PLE

Letting the following:

f(d) =1

darctan

50

d

0.0138048

f 0(d) =1

d

1

1 + 250d2

The root is near d = 55, so subsituting values for f(d) and f 0(d) into the Newton-Raphson iterationdn+1 = dn f(dn)/f 0(dn) gives:

n dn0 551 54.083573912 54.077016153 54.071397684 54.064979065 54.064918526 54.064918527 54.064918528 54.064918529 54.0649185210 54.06491852

Resulting in our final solution that:

v = 54.065tan(1.2128 0.04665t) (11)

7.2 Conclusion Comparing New Predictions and Data

The new data much more closely matches with our expected results, in fact the graph shows them almostindistinguishable from each other.

The data sets are below:

t (s) 0 1 2 3 4 5 6 7 8 9vactual (m/s) 96 89 82 77 72 68 64 61 58 55vpredicted (m/s) 96 89 82 77 72 68 64 61 58 55

t (s) 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26vactual (m/s) 50 46 41 38 34 31 27 24 21 18 16 13 10 8 5 3 0vpredicted (m/s) 50 48 41 38 34 31 27 24 21 18 17 14 10 8 5 4 0

10

EXAM

PLE

0 2 4 6 8 10 12 14 16 18 20 22 24 260

10

20

30

40

50

60

70

80

90

100

time (s)

speed(m

/s)

11

EXAM

PLE

7.3 Recommending minimum runway length

The minimum length of the runway for the aeroplane to safely land must be the area under the finalspeed / time graph (on page 11).

Our two final equations are:

v0to9.5 =ds

dt=

1

0.0008628t+ 0.01042

and

v9.5to26 =ds

dt= 54.065tan(1.2128 0.04665t)

Hence the distance covered during landing, s is the integral of these velocities.

i) From t=0 to t=9.5:

s1 =

Z 9.50

1

8.628 104t+ 0.01042 dt =1

8.628 104Z 9.50

8.628 1048.628 104t+ 0.01042 dt

=) s1 = 1209.5ln(0.0008268t+ 0.01042)9.50

= 4818.28 (5520.19) = 701.9m

ii) For t=9.5 to t=26, we have:

s2 =

Z 269.5

54.065tan(1.2128 0.04665t) dt

=) s2 = 1159.9 lnsec(1.2128 0.04665t)269.5

= 0 (383.9) = 383.9m

Hence the total length used during landing 383.9 + 701.9 = 1085.88m

A sensible recommendation for the minimum runway length would be to round up to 1.1km, consideringthe vast number of assumptions about the conditions made initially for our model.

12

EXAM

PLE

IntroductionSimplifying and Setting up the ModelAssumptions MadeRelevance of AssumptionsEstablishing Differential Equations

Manipulating the ModelSolving the Differential EquationsChoosing the parameters of the DEParticular Solutions to the Differential EquationsPredictions Produced from Equations

Verifying the ModelCollecting the dataThe Data

Comparison of Predictions and DataComparisonVariations of Parameters

Revision of the ModelAmmendmentsEstablishing New Equations

Assesment and ConclusionsSolving the New EquationsConclusion Comparing New Predictions and DataRecommending minimum runway length