Mathematical modeling of pathway of a plane

I have chosen mathematical modeling of the pathway of a plane as the topic of this project. I chose this topic as I always wondered how an aircraft floats in air and what was more fascinating to me was how it climbs up and moves without falling down. My curiosity intrigued me to find out the reason for floating and to understand the motion of an aircraft. I went through many websites and educational content online and after my preliminary research; I found out that different types of aircrafts have different flight motion and mechanics for example; the motion and flight mechanics of a rocket which goes up into the space is different to that of a jet or a passenger airplane. In my higher studies, I got introduced to motion in a plane and there I learnt about projectile motion and how effectively mathematics is used to analyze the motion of an object (in air). It gave me the slightest idea of how mathematics could be used to describe the motion of an aircraft and gave me a different perspective of imagining things. I have tried explaining mathematically the motion of a plane in this project with whatever expertise I have gained in mathematical modeling from projectile motion and motion of an object in a plane in general. However, in order to fully explain the motion of an aircraft analytically, some higher level mathematical tools such as three dimensional linear algebra which includes Euclidean angles, Euclidean space and the concept of aerodynamics needs to be studied, which is beyond my scope and the scope of the syllabus.

Mathematical modeling of pathway of a plane. To derive an equation of motion, the clear understanding of the forces acting on the aircraft is essential. To begin with the forces acting on the aircraft in case of multiple situations steady flight, rotation and curvilinear motion will be derived. Vector mathematics will be used to split them into components and thus derive an equation of motion. Following this, a special case steady flight was taken to analyze the equation of motion.

To describe the motion of the aircraft, I have used mainly two mathematical tools.

Vectors are physical quantities which have both magnitude and direction for example;

 

velocity, acceleration, force etc.

Let two vectors \(\overrightarrow A\) and \(\overrightarrow B\) act at some angle to each other as shown in Fig. (a). Coinciding the tail of the vector \(\overrightarrow A\) on the head of vector \(\overrightarrow B\), we get the resultant vector \(\overrightarrow R\) as shown in Fig. (b). The vectors \(\overrightarrow A\) and \(\overrightarrow B\) are represented by the sides\(\overrightarrow {OP}\) and \(\overrightarrow {PQ}\) of the triangle OPQ.

 

Vector addition and resolution of vectors can be used to resolve a force vector into its vertical and horizontal
components

In the above figure, \(\overrightarrow F\) is the force vector which makes an angle θ with respect to the X axis. This force \(\overrightarrow F\) can be resolved into its two rectangular components using basic trigonometry. The vertical component is F sin sin θ and the horizontal component is F cos cos θ as shown in the above figure.

Let y = f (x) be a single valued function of x defined in some interval. Let x be any value of x in the domain of definition of the function and the corresponding value of y is y = f (x). Suppose, for an increment ∆x of x the corresponding increment in y is ∆y.

 

Then \(\frac{∆y}{∆x}=\frac{∆y}{∆x}\) is called the derivative of the function y with respect to x, provided the limit exists. In this project, we do not require a good understanding of the concept of integration. We just require the skill to solve some basic functions to integrate. Integration may be referred to as antiderivative where we find the original function which was differentiated with respect to some another variable.

The four forces acting on the aeroplane (as shown in the above figure) flight lift, drag, thrust and weight are denoted by L, D, T and W respectively. The free stream velocity is denoted by \(V_∞\) and the flight path is denoted by S. The free stream velocity acts in the direction of the flight of the plane.

 

By definition, lift L and drag D are perpendicular and parallel to the free stream velocity \(V_∞\). Lift and drag are two aerodynamic forces. The weight W always acts towards the centre of the Earth. The thrust T does not necessarily act opposite to the direction of the free stream velocity. T can be inclined at an angle ∈ relative to the flight path.

As shown in the above figure, the aeroplane is climbing along a curved flight path which makes an instantaneous angle θ with respect to the horizontal. Thus, the free stream velocity V_∞ makes an angle θ with respect to the horizontal. This angle is called the local climb angle of the aeroplane.

 

Lift and drag act perpendicular and parallel to V_∞. The weight of the aeroplane is perpendicular to the Earth’s surface. When the aeroplane is climbing, the weight W of the plane is inclined at an angle θ relative to the lift.

In the figure below, the aeroplane rotates by an angle about its longitudinal axis. The plane of symmetry changes from the plane of the paper to the plane shown in the figure below (front view of the aeroplane).

Fig. (7) is the front view of the aeroplane and the forces acting on it on plane of symmetry AA which is perpendicular to the free stream velocity V_∞. The plane of symmetry is inclined at an angle ϕ with respect to the vertical.

The equations of motion of an aircraft can be derived using Newton’s law of motion F = ma, where F denotes force on an object; m is mass of the object and acceleration of the object. Since we are concerned about only the translational motion of the aircraft, we can consider it to be a point mass4 at its centre of gravity.

The forces drag D and weight W are on the plane of the paper. The component of the flight on this plane is L cos cos θ . The radius of curvature is r₁.

 

Then, the component of the force parallel to the flight path of the plane is,

 

F_parallel = T cos cos ∈ - D - W - sin sin θ

 

=> m \(\frac{dV_∞}{dt}\) = T cos cos ∈ - D - W sin sin θ...(i)

 

where \(\frac{dV_∞}{dt}\) is the rate of change of free stream velocity with respect to time.

 

The component of the force acting perpendicular to the direction of the flight path is,

 

F_{perpendicular} = Lcosϕ + Sin ∈  cosϕ - W cosθ

 

The radial acceleration due to the curvilinear motion, perpendicular to the flight is,

 

 \(ar = \frac{V_∞}{r_1}\)

 

Therefore using this expression of the radial acceleration we get,

 

\(m\frac{V_∞^2}{r_1}\) = Lcosϕ + T sin ∈ cosϕ - W cosθ...(ii)

 

Here, we get our first two equations of motion. Now, the motion of the flight is analyzed from the top view. According to the above diagram, the component of the force acting on the aeroplane is,

 

F₂ = Lcosϕ + T sin∈  sinϕ

 

The radial acceleration due to the curvilinear motion6 is,

 

\(a_2=\frac{(V_∞cosθ)(V_∞cosθ)}{r_2}\)

 

Therefore, the equation changes to,

 

\(m=\frac{(V_∞cosθ)(V_∞cosθ)}{r_2}\) = Lcosϕ + T sin∈  sinϕ...(iii)

 

From equation (i), we can calculate the free stream velocity using integration and other values provided. For example, if we have climb angle θ = 30°, weight of the aircraft W = 900N, thrust of the aircraft T = 800N, drag D = 500N, mass m = 40000kg, time t is from 0sec to 5 secs, instantaneous angle ∈ = 5° and angle of rotation ϕ about the longitudinal axis is 30°, we calculate the free stream velocity from equation (I).

 

\(m\frac{dV_∞}{dt}\) = T cos cos θ - D - W sin sin θ

 

=> m dV_∞ = (T cos cos θ - D - W sin sin θ)dt

 

=> m dV_∞ = \(\frac{(T\ cos\ cos\ θ-D-W\ sin\ sin\ θ)dt}{m}\)

 

\(=>\displaystyle\int\limits^5_0■dV_∞=\displaystyle\int\limits^5_0■\frac{(T\ cos\ cos\ θ-D-W\ sin\ sin\ θ)dt}{m}\)

 

=> V_∞ = \(\frac{1}{m}\displaystyle\int\limits^5_0■\) (T cos cos θ - D - W sin sin θ)dt

 

=> V_∞ = \(\frac{1}{40000}\displaystyle\int\limits^5_0■\)(800 cos cos 30° - 500 - 1000 sin sin 30°)dt

 

=> V_∞ = \(\frac{1}{40000}\displaystyle\int\limits^5_0■\) - 307.17 dt

 

=> V_∞ = \(\frac{1}{40000}\) (-1535) km/sec

 

=> V_∞ = - 0.038375 km/sec

 

∴ V_∞ ≈ 0.038 km/sec

 

(neglecting the direction of the freestream velocity)

 

Using equation(ii), we can find out the radius of curvature if the free stream velocity is known and other data is provided. Considering the value of V_∞ which we got, lift of the aircraft L to be 700N and the data which we have taken earlier we get,

 

m \(\frac{V_∞\ ^2}{r_1}\) = L cosϕ + T sin∈  cosϕ - W cosθ

 

=> 40000 \(\frac{0.038^2}{r_1}\) = 700 cos cos 30° + 800 sin sin 5° cos cos 30° - 40000 cos cos 30°

 

=> r₁ = 40000 × 0.038 × 0.038 × \(\frac{1}{-33974.41}\) km

 

∴ r₁ ≈ 0.0447km (neglecting the direction of the radius of curvature)

 

Similarly, using equation(iii), by putting the values in the equation.

 

m \(\frac{(V_∞cosθ)(V_∞cosθ)}{r_2}\) = Lsinϕ + Tsin∈ sinϕ

 

=> 40000 \(\frac{0.038\ cos\ cos\ 30°×0.038\ cos\ cos\ 30°}{r_2}\) = 700 sin sin 30° + 800 sin sin 5° sin sin 30°

 

=> r₂ = \(\frac{43.32}{384.86}\) km

 

=> r₂ ≈ 0.112km

 

Thus, 0.112km is the radius of curvature is r₂.

For a steady flight, the climb angle θ and the angle of rotation ϕ about the longitudinal axis of the aircraft are zero. Since for a steady flight, the flight path is straight, the radius of curvature r is infinitely large hence, \(\frac{V^2}{r}\)) is zero.

 

Therefore, the resulting equations are:

 

m\(\frac{dV_∞}{dt}\) = T cos cos ∈ - D - W sin sin θ

 

=> 0 = T cos cos ∈  - D

 

And,

 

m\(\frac{dV_∞\ ^2}{r_1}\) = Lcosϕ + T sine cosϕ - Wcosθ

 

=> 0 = L + T sin sin  ∈ - W

 

If we neglect the instantaneous angle ∈ then we get,

 

T = D and L = W.