Project-1: Powertrain for aircraft in runways
| Aircraft Types | Maximum Take-Off Weight (kg) MTOW |
| Antonov An-225 | 640,000 |
| Airbus A380-800 | 575,000 |
| Boeing 747-8F | 447,700 |
| Boeing 747-400 | 396,900 |
| Airbus A340-500 | 371,950 |
| McDonnell Douglas MD-11 | 273,300 |
| Boeing 787-10 | 254,000 |
| Airbus A340-200 | 253,500 |
| Ilyushin IL-96-300 | 250,000 |


Carcass plies are used to form the tire. They are sometimes called casing plies. An aircraft tyre is constructed for the purpose it serves.
1. They support the weight of an aircraft while it is on the ground.
2. It also provides the necessary traction for braking and stopping of an airplane.
3. Tyre also helps to absorb the shock of landing and provide cushioning the roughness of takeoff, roll-out, and taxi operations.
Unlike an automobile or truck tire, it does not have to carry a load for a long period of continuous operation. However, an aircraft tyre absorbs the high impact loads of landing, and also it’s operating at high speeds for a short time when required.
Tubeless tires are more advantageous over tube-type. There is no longer the use of tube-type tire in recent aviation. Nowadays all airliners are using tubeless tires. Tubeless that are meant to be used without a tube has the word TUBELESS on the sidewall of the tyre.
Below are the tire building terminologies.

The coefficient of rolling resistance (`murr`) depends on many factors like the Tyre Pressure,Tyre Temperature, Material of the Tyre, Thread pattern of the Tyre, and also the Road Roughness and Liquids on the Road. These factors determine the rolling resistance force.
to 320 psi. It's really pressurized air that's so strong.
Like everything in aviation, tires must meet specific and demanding rules—for example, a tire must withstand four times its rated pressure for at least three seconds.
Q. 6) With necessary assumptions, calculate the force and power required to push/pull an aircraft by a towing vehicle.
1) Forces acting on the Aircraft and Tug (Towing vehicle):-
There are mainly two forces acting in the Towing condition;
A) Rolling resistance force:
The Rolling resistance force is experienced by the tyres of the vehicle.
The rolling resistance force is given by the expression,
`Frr = murr * m*g`
Where,
`m`= mass of the vehicle (kg)
`g`= gravity ( 9.81 `m/s^2`)
`murr`= coefficient of rolling resistance.
Calculations:
1) For Aircraft:
Let us assume,
Weight of the Aircraft = 500000 Kg
Coefficient Rolling Resistance for an Aircraft (`murr`) = 0.004
Hence,
Rolling resistance Force of Aircraft = `murr*m*g` = `0.004 * (500000)*(9.81)`
= `19.620` kN
2) For Towing vehicle :
Let us assume,
Weight of the Towing tractor = 50000 kg
Coefficient of Rolling Resistance of Towing vehicle (`mu_(rr)`) = 0.001
Hence,
Rolling resistance force of towing vehicle = `murr*m*g` = `0.001 * (50000)*(9.81)`
Rolling resistance force of towing vehicle = `0.49` KN
Now,
Total Rolling Resistance Force = Rolling Resistance Force of aircraft + Rolling Resistance Force Towing tractor
= `19.620 + 0.49`
Total Rolling Resistance Force = `20.11` kN
B) Air Drag Force:-
The air drag force is the force experienced by the vehicle due to the air striking when moving.
The Air Drag Force is given by;
`Fad = 1/2*rho*A*v^2*C_d`
Where,
`rho` = density of air
`A`= frontal area of the vehicle on which the air strikes
`v`= velocity of the vehicle
`C_d`= Drag Coefficient
Calculations:
Here,
Density of Air (rho) = 1.25 kg/m^3
Let us assume maximum speed,
The velocity of the Aircraft while Towing = 10 km/h = 2.77 m/s ( Towing speed is in between 5 km/h to 10 km/h and it does not exceed 10 km/h in any case )
1) For Aircraft:
Assume,
The frontal area of the Aircraft = 20 `m^2`
The coefficient of Drag of the Aircraft = 0.2
`therefore`Air drag force = `1/2* rho * A * v^2 * C_d`
= 0.5 * 1.25 * 20 * (`2.77^2`) * 0.2
= 0.019 kN
2) For Towing Vehicle:
Assume,
The frontal area of the Towing vehicle = 2 `m^2`
The coefficient of Drag of the Towing vehicle = 1.1
`therefore`Air drag force = `1/2* rho * A * v^2 * C_d`
= 0.5 * 1.25 * 2 * (`2.77^2`) * 1.1
= 0.011kN
`therefore` The total Air Drag force = Air Drag force of the aircraft + Air Drag force of the towing vehicle
= 0.019 + 0.011 = 0.03 kN
Hence,
The Total Force that the towing vehicle needs to overcome in order to pull the Aircraft = Total Rolling Resistance Force + The total Air Drag force
= 20.11 + 0.03 = 20.14 kN
`therefore` Total Force reqiured is 20.14 kN.
2) Power Required:-
The maximum speed of Pull = 10 km/h = 2.77 m/s
Power required by the towing vehicle = Total Forces * Velocity = 20.14 kN * 2.77 m/s
Power required by the Towing vehicle = 55.78 kW.
Q. 7) Design an electric powertrain with the type of motor, it’s power rating, and energy required to fulfill aircraft towing application. Estimate the duty cycle range to control the aircraft speed from zero to highest. Make all required assumptions. Prepare a table of assumed parameters. Draw a block diagram of the powertrain.
| Sr. No. | Assumed Parameters | Values |
| 1. | Weight of the Aircraft | 500000 kg |
| 2. | Weight of the Towing Tractor | 50000 kg |
| 3. | Rolling resistance coefficient of aircraft | 0.004 |
| 4. | Rolling resistance coefficient of towing tractor | 0.001 |
| 5. | The velocity of the Aircraft while Towing | 2.77 `m/s^2` |
| 6. | Density of Air | 1.25 `(kg)/m^3` |
| 7. | The Frontal area of the Aircraft | 20 `m^2` |
| 8. | The Frontal area of the towing vehicle | 2`m^2` |
| 9. | The coefficient of Drag of the Aircraft | 0.2 |
| 10. | The coefficient of Drag of the towing vehicle | 1.1 |
| 11. | Operational time of the Towing for one cycle | 0.5 hours |
With these assumed parameters we have calculated the power required for towing. Hence, Power = 55.78 kW.
Energy Required:
Required Energy(E) for towing application based on the assumed time,
Energy =`P * t`
= 55.78 * 0.5
= 27.89 kWh
Hence we will use a battery that stores more than 27.89 kWh energy.
We are using 'NetGain HyPer9 HV' AC Motor
Its maximum power is 88 kW
Let's assume continuous power is 80 kW
Hence,
Duty cycle =`(55.78/80)*100` = 80 %
80% will be used for a half-hour to complete the task in one cycle from zero to high.
Block diagram for the electric towing vehicle.
CONCLUSION:-
Weights of different aircraft are enlisted. The difference between airspeed and ground speed is explained. The reasons for not recommending the use of aircraft engine power is elaborated well. The force and power requirement is calculated for towing vehicles for pull purposes. And also electric drivetrain is designed for towing purposes.