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Essay: Aircraft – mathematics

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Math SL
Internal Assessment
Lift and Drag
Introduction
When you look at aircrafts, they look like they shouldn’t be able to leave the ground because of how big they are. I always watched aircrafts, take off and land, over and over. According to Newton’s Third Law, every action has an equal and opposite reaction, lift opposes weight and enables flight. This interest gradually grew, making me use the concept of Lift and Drag as an exploration for my Math IA.
Aircrafts come in a variety of shapes and sizes, they are heavily engineered pieces of work that allow easy global transportation and lets us see the world. The amount of lift generated by an aircraft depends on the size of the aircraft. Lift is an aerodynamic force and therefore depends on the pressure variation of the air around the body as it moves through the air. The total aerodynamic force is equal to the pressure times the surface area around the body.1 Like the other aerodynamic force, drag that acts in the direction opposite to the motion of the aircraft, the lift is directly proportional to the area of the aircraft. Doubling the area, hence doubles the lift. When an airplane takes off, the pilot applies as much thrust as possible to make the airplane roll along the runway. But just before lifting off, the pilot “rotates” the aircraft. The nose of the airplane rises, thus increasing the angle of attack (the inclination of the aircraft) and producing the increased lift needed for takeoff.
Generation of Lift is similar to that of drag, the same principle of “no motion no drag” and “no fluid no drag” applies to lift as well, as its generated by the physical contact between solid body and fluid and not by forces caused due to electromagnetic field or magnetic field or any force field in that matter.
On the other hand, Drag, an aerodynamic mechanical force, continuously opposing the aircraft’s motion in air, is generated by each and every part of an aircraft, including the engines. In order to generate drag, a physical contact is necessary between the solid body and a fluid (gas or liquid). Forces like the ones existing in magnetic field or gravitational field cannot help generate drag, as there’s no contact between the solid and fluid.
Factors affecting both lift and drag are in the majority, which means generation of either lift or drag will eventually result in the generation of the other depending on the additional factors required for the same. Drag can be explained in many ways, as aerodynamic friction and aerodynamic resistance (also called form drag).
Friction
Generation of lift forms a component of drag, known as induced drag, which occurs due to the spanwise distorted flow around the wing tips which is also a result of pressure variance from top to bottom of the wing.
All the factors mentioned above, for both lift and drag, can be combined in order to form equations for both quantities, and these equations are called lift equation and drag equation respectively.
Existing from the year 1900 onwards, these equations have two versions to it, one is an older equation which was used at the time of the Wright brothers and another is the modern equation is is currently used to calculate lift and drag for aircrafts with different features such as shape, thrust, weight etc.
I will be using both the equations and compare the lift and drag obtained respectively for different aircrafts.
Lift Equation of the Early 1900’s
From 1900 onwards, the Wright brothers started designing gliders and aircrafts. In order to achieve this, size of wings were crucial for which they required good knowledge of maths and physics.
They ran a bicycle shop and were well aware about newton’s laws of motion, force and torque.
Square of velocity, was something already determined to be a factor on which lift depends, when wrights started their studies, it also varied linearly along with the surface area of the object.
Smeaton’s coefficient was used to characterize dependence on properties of air on one foot sq. flat plane moving at 1 mile per hour. Early aerodynamicists concluded that some portion of pressure force was converted into lift for any object moving in air.
This was represented by the coefficient.
The final equation was:
L = k * V^2 * A * cl …..
where:
L is the lift
k is the Smeaton coefficient
V is the velocity
A is the wing area
cl is the lift coefficient.
The above equation differs from the modern equation drastically. Dynamic air of moving air is used for pressure dependence whereas Smeaton’s coefficient is used in this.
The modern coefficient relates the lift force on the object to the force generated by dynamic pressure times area. While the 1900’s equation relates the lift force to the drag of equal plate area.
Modern Lift Equation
As we know the complexity behind the dependence on inclination, compressibility, body shape and air viscosity, using a single variable to characterize the dependency is the most efficient way to find lift force and is the same concept used in the modern lift equation. “Cl” is the variable assigned for the coefficient. The value of Cl, that is used to determine lift, inclination of object, shape and air conditions are required. Taking general examples, where for thin airfoils that have small angles of attack, 2 times pi (3.14159) times angle of attack, is the lift coefficient which is measured in radians.
Cl = 2 * pi * angle (in radians)
The modern lift equation states that lift is equal to the lift coefficient (Cl) times the density of the air (r) times half of the square of the velocity (V) times the wing area (A).
L = .5 * Cl * r * V^2 * A …..
With the modern lift equation one can calculate the amount of lift force produced at a certain velocity for the respective area, or calculate the wing size, to lift any amount of weight, for a given velocity.
Drag Equation of the Early 1900’s
As I mentioned earlier, the same parameters and the same factors affect drag force. Just like how the early aerodynamicists, used Smeaton’s Coefficient instead of dynamic pressure to calculate the drag of any object moving through air.
The equation they resulted in was:
D = k * V^2 * A * cd ……
where:
D is the drag
k is the Smeaton coefficient
V is the velocity
A is the wing area
cd is the drag coefficient.
Similar to lift equations, the above drag equation differs from modern drag equation as Smeaton’s Coefficient for pressure dependence is used here whereas in the modern drag equation it is seen that dynamic pressure is used instead.
Modern Drag Equation
Now, in the modern drag equation, factors like viscosity, compressibility of air, density of air and velocity between object and air is included to find the drag that is generated for any object moving through air.
The equation is given by:
D = .5 * Cd * r * V^2 * A ….
where:
drag is (D)
drag coefficient is (Cd)
density of the air is (r)
velocity is (V)
wing area (A).
While we use quantities like viscosity, compressibilities and velocity, we must understand the complexity behind it. “Cd” the coefficient of drag has an essential calculation to itself as it requires the sum of drag at zero lift and something called induced drag which I have defined earlier in the introduction of drag.
The equation for coefficient turns out to be:
Cd = Cd0 + Cl^2 / ( pi * Ar * e) …..
where:
Cd is drag coefficient
Cd0 is the drag coefficient at zero lift
Cl is lift coefficient
Ar is aspect ratio
e is efficiency factor.
Comparison of Lift equations
Calculation using early Lift equation
Here I will consider two airbuses, the Boeing 747-8 intercontinental airbus and Airbus A380,
assuming both are empty to ease calculation and find the lift force using the early lift equation for both the cases and plotting an airspeed vs lift whereas airspeed vs net force graph for each.
Boeing 747-8 Intercontinental
Max Cruise Speed
917 km/h
Empty Weight
213 kg or 2088.817N
Wing Span
68.45 m
Wing Area
554 m2
Length
76.25 m
Height
19.36 m
Coefficient of Lift
0.52
Smeaton’s coefficient
0.005
Data from Boeing
Above we have the required factors to calculate the lift force using the lift equation:
L = k * V^2 * A * Cl…….(1)
Given:
k=0.005
V=917 km/h
A= 554 m2
Cl=0.52
Substituting the above values in equation (1), we get
L= 0.005*(917)^2*554*0.52
∴ L= 1,207,253.9752 N
∴ Net force = 1,207,253.9752 – 2088.817 = 1,205,165.16 N
Airbus A380
Max Cruise Speed
945 km/h
Empty Weight
277 kg or 2716.443 N
Wing Span
79.80 m
Wing Area
845 m2
Length
73 m
Height
24.10 m
Coefficient of Lift
2.74
Smeaton’s coefficient
0.005
Data from Airbus
Above we have the required factors to calculate the lift force using the lift equation:
L = k * V^2 * A * Cl…….(1)
Given:
k=0.005
V=945 km/h
A= 845 m^2
Cl=2.74
Substituting the above values in equation (1), we get:
L=0.005*(945)^2*845 *2.74
∴ L= 10,338,103.9125 N
∴ Net force = 10,338,103.9125 – 2716.443 = 10,335,387.47 N
Calculations using Modern Lift Equation
Similar to the previous calculations, here I will consider the same two airbuses, the Boeing 747-8 intercontinental airbus and Airbus A380, assuming both are empty to ease calculation and find the lift force using the modern lift equation for both the cases and plotting airspeed vs net force graph for each.
Boeing 747-8 Intercontinental
Max Cruise Speed
917 km/h
Empty Weight
213 kg or 2088.817N
Wing Span
68.45 m
Wing Area
554 m2
Length
76.25 m
Height
19.36 m
Coefficient of Lift
0.52
Density of air (r)
0.30267 kg/m³
Data from Boeing
Above we have the required factors to calculate the lift force using the lift equation:
L = .5 * Cl * r * V^2 * A ….(1)
Given:
V=917 km/h
A= 554 m2
Cl=0.52
r=0.30267 kg/m³
Substituting the above values in equation (1), we get:
L= 0.5*0.52*0.30267*(917)^2*554
∴ L= 36,659,890.27766 N
∴ Net force = 36,659,890.27766 – 2088.817 = 36,657,801.46 N
Airbus A380
Max Cruise Speed
945 km/h
Empty Weight
277 kg or 2716.443 N
Wing Span
79.80 m
Wing Area
845 m2
Length
73 m
Height
24.10 m
Coefficient of Lift
2.74
Density of air (r)
0.30267 kg/m³
Data from Airbus
Above we have the required factors to calculate the lift force using the lift equation:
L = .5 * Cl * r * V^2 * A ….(1)
Given:
V=945 km/h
A= 845 m^2
Cl=2.74
r=0.30267 kg/m³
Substituting the above values in equation (1), we get:
L=0.5*2.74*(945)^2*845*0.30267
∴ L= 312,903,391.1196 N
∴ Net force = 312,903,391.1196 – 2716.443= 312,900,674.68 N
1,2 https://www.grc.nasa.gov/www/k-12/airplane/size.html
Comparison of Drag equations
Calculation using early Drag equation
Here I will consider two airbuses, the Boeing 747-8 intercontinental airbus and Airbus A380,
assuming both are empty to ease calculation and find the lift force using the early drag equation for both the cases.
Boeing 747-8 Intercontinental
Max Cruise Speed
917 km/h
Empty Weight
213 kg or 2088.817N
Wing Span
68.45 m
Wing Area
554 m2
Coefficient of Drag
0.024
Smeaton’s coefficient
0.005
Data from Boeing
Above we have the required factors to calculate the drag force using the drag equation:
D = k * V^2 * A * cd ……(1)
Where,
k=0.005
V=917 km/h
A=554 m2
cd=0.024
Substituting values in the equation (1), we get,
D=0.005*(917)^2*554*0.024
∴D=11,180,521.10616 N
Airbus A380
Max Cruise Speed
945 km/h
Empty Weight
277 kg or 2716.443 N
Wing Area
845 m2
Coefficient of Drag
0.0265
Density of air (r)
0.30267 kg/m³
Smeaton’s coefficient
0.005
Data from Airbus
Above we have the required factors to calculate the drag force using the drag equation:
D = k * V^2 * A * cd ……(2)
Where,
k=0.005
V=945 km/h
A=845m2
cd=0.0265
Substituting the above values in equation 2, we get,
D= 0.005*(945)^2*845*0.0265
∴D= 99,985.3115625 N
Calculations using Modern Drag Equation
Similarly, here I will consider the same two airbuses, the Boeing 747-8 intercontinental airbus and Airbus A380,
assuming both are empty to ease calculation and find the drag force using the modern drag equation in both the cases.
Boeing 747-8 Intercontinental
Max Cruise Speed
917 km/h
Empty Weight
213 kg or 2088.817N
Wing Area
554 m2
Coefficient of drag
0.024
Density of air (r)
0.30267 kg/m³
Data from Boeing
Above we have the required factors to calculate the drag force using the drag equation:
D = .5 * Cd * r * V^2 * A ….(1)
Where,
Cd= 0.024
r=0.30267 kg/m³
A= 554 m2
V=917 km/h
Substituting the above values in equation 1, we get,
D= 0.5*0.024*0.30267*(917)^2*554
∴D=1,691,994.94 N
Airbus A380
Max Cruise Speed
945 km/h
Empty Weight
277 kg or 2716.443 N
Wing Area
845 m2
Coefficient of Drag
0.0265
Density of air (r)
0.30267 kg/m³
Data from Airbus
Above we have the required factors to calculate the drag force using the drag equation:
D = .5 * Cd * r * V^2 * A ….(2)
Where,
Cd= 0.0265
r=0.30267 kg/m³
A= 845 m2
V=945 km/h
Substituting the above values in equation 2, we get,
D= 0.5*0.0265*0.30267*845*(945)^2
∴D= 302,625.542 N
The lift and drag I’ve worked out above are for ideal condition. This means that we aren’t considering the wind velocity as it varies from place to place, causing a complexion when it comes to finding lift and drag.
This is when turbulence, cross-winds and other natural phenomenons affects the flying of an aircraft which just means that the lift force has an opposing force acting towards it, causing unbalanced landings and take-offs.
One of the hardest part about the whole ia would be the graph comparison for the early and modern lift calculations, as the values for both are nearly a 7 digit figure, finding out the instantaneous lift using the graph is very complex hence formulas make it much easier and less time taking for an individual in such cases.

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