Project-1: Powertrain for aircraft in runways
PROJECT : POWERTRAIN FOR AIRCRAFT IN RUNWAYS
AIM OF THE PROJECT:
- Detailed overview of air-craft and its parts.
- Modelling and estimation of necessary parameters required to control the motion of aircraft in runways.
- Observing the behaviour of aircrafts motion in grounf and during take-off.
Objective:
- 1. Search and list out the total weight of various types of aircrafts.
- 2. Is there any difference between ground speed and air speed?
- 3. Why is it not recommended to use aircraft engine power to move it on the ground at Airport?
- 4. How an aircraft is pushed to runway when its ready to take off?
- 5. Learn about take off power, tyre design, rolling resistance, tyre pressure, brake forces when landing.
- 6. A. With necessary assumptions, calculate the force and power required to push / pull an aircraft by a towing vehicle.
- B. Develop the model for the calculated force and power using Simulink.
- 7. A. Design an electric powertrain with type of motor, it’s power rating, and energy requirement to fulfill aircraft towing application in Simulink. 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 powertrain.
- (Hint :DC7 Block)
- B. Also, Design the parameters in excel sheet.
Solution:
STUDY 1: As we know that ,weight is the force generated by the gravitional attraction of the earth which we can estimate by the given formula w=m*g, where w=is the weight, m is the mass, g is the gravitational constants g=9.8m/sec^2 or 32.2ft/sec^2 . In case of a whole aircraft there are different parts and each part of the aircraft bears a unique weight and mass example : the wings, the engines, the payload, and the fuel.The mass of the individual components can be calculated if we know the size of the components and its chemical composition example: iron, plastic, aluminium, gasoline etc. Therefore, the total weight of the aircraft is the sum of the individual components given as: W=w(Fuselage)+w(Payload)+w(Engines)+w(Wings)+....
It can be pictured as :-

Center of gravity (CG) of an aircraft is calculated as follows:
- Determine the weights and arms of all mass within the aircraft.
- Multiply weights by arms for all mass to calculate moments.
- Add the moments of all mass together.
- Divide the total moment by the total mass of the aircraft to give an overall arm.
The arm that results from this calculation must be within the center of gravity limits dictated by the aircraft manufacturer. If it is not, weight in the aircraft must be removed, added (rarely), or redistributed until the center of gravity falls within the prescribed limits.
Aircraft center of gravity calculations are only performed along a single axis from the zero point of the reference datum that represents the longitudinal axis of the aircraft (to calculate fore-to-aft balance). Some helicopter types utilize lateral CG limits as well as longitudinal limits. Operation of such helicopters requires calculating CG along two axes: one calculation for longitudinal CG (fore-to-aft balance) and another calculation for lateral CG (left-to-right balance).
The weight, moment and arm values of fixed items on the aircraft (i.e. engines, wings, electronic components) do not change and are provided by the manufacturer on the Aircraft Equipment List. The manufacturer also provides information facilitating the calculation of moments for fuel loads. Removable weight items (i.e. crew members, passengers, baggage) must be properly accounted for in the weight and CG calculation by the aircraft operator.
| Mass (lb) | Arm (in) | Moment (lb-in) | |
|---|---|---|---|
| Empty aircraft | 1,495.0 | 101.4 | 151,593.0 |
| Pilot and passengers | 380.0 | 64.0 | 24,320.0 |
| Fuel (30 gallons @ 6 lb/gal) | 180.0 | 96.0 | 17,280.0 |
| Totals | 2,055.0 | 94.0 | 193,193.0 |
To find the center of gravity, we divide the total moment by the total mass: 193,193 / 2,055 = 94.01 inches behind the datum plane.
Aircraft authorised gross weight is defined as the total aircraft weight at any moment during the flight or ground operation. The authorised weight limits that can be legally be used by any operator or airline are listed in the AFM and the weight and balance manual. They are also referred as purchased weights.
The classification of different types of weight under gross weight can be made in the following categories:
- Maximum ramp weight(MRW):- It is defined as the maximum weight authorised for taxiing or towing on the ground as limited by aircraft strength or airworthiness requirements. The difference between the maximum ramp weight and the maximum take-off weight depends on the size of the air-craft, the number of engines , APU operation and engine fuel consumption. Which is typically assumed for 10-15 minuts of taxiing, towing of run-up operations.
- Maximum take-off weight(MTOW):- It is defined as the maximum certified design weight for take-off run as limited by air-craft strength and aieworthiness requirement. the maximum take-off weight is always less than the maximum ramp weight to allow fuel burned during taxi by the engines and the APU.
- Maximum landing weight(MLW):- It is defined as the maximum design certified weight for normal landing of the aircraft. It mainly depends on the landing gear strength or the landing impact loads on certain parts of the wing and chassis structure.
- MDLW<MDTOW<MRW
- Maximum zero fuel weight(MZFW):- The maximum zero fuel weight is the maximum certified of the aircraft without the usable fuel and others limited by aircraft strength and airworthiness requirements. In other word it is the maximum weight permitted before the usable fluid and other usable items are loaded in the aircraft.
- Operational empth weight(OEW):- The OEW is fixed for a specific aircraft and only gets changed during operational changes and maintenance. OEW is the empty weight that is used in flight simulator or usuall planning software.
- Actual zero fuel weight(AZFW):- If we add the payload to the Empty plane weight then we get the actuall zero fuel weight. Basically the payload includes the passenger luggage weight and cargo weight given by the following relationship.
- Minimum Flight weight(MFW):- Minimum certified weight for flight as limited by aircraft strength and airworthiness requirements. The total weight of an aircraft includes aircrafts's empth weight +payload.
- Aircrafts's empty weight:- The aircraft weight with fuel left in tank +engine oil +other operating fluids
- Payload:- The fuel weight+ passengers weight +other operating fluids.
| Operational empty weight(OEW) |
+
| Payload |
=
| Actuall zero fuel weight(AFZW) |
| Aircraft type | Maximum take-off weight(kg) | Maximum fuel(kg) |
| Boeing707-320 | 151500 | 72503 |
| Boeing747-200 | 52610 |
14520 |
| AirbusA300-B4 | 165000 |
46654 |
| Boeing757-200 | 115650 | 36260 |
| DouglasDC-8-32 | 140000 | 31020 |
| Boeing767-400 | 209111 | 73363 |
| Boing737-900 | 79016 |
20894 |
| AirbusA340-600 | 368000 | 156487 |
| The embraer phenom 300 | 8000 | 1545 |
STUDY 2:> AIR-SPEED Vs GROUND- SPEED
Ground speed: It is the horizontal speed of an aircraft relative to the ground. An aircraft heading vertically would have a ground speed of zero. Ground speed can be determined by the vector sum of the aircraft's true airspeed and the wind speed and the direction of motion.
Wind speed: If we pick a reference point on the ground, the air moves relative to the reference point at the wind speed. A positive velocity is defined in the direction od the aircraft's motion.
Airspeed: The important part in the generation of lift of aircraft is the relative velocity between the object and the air, which is called air-speed. It cannot be directly measured from the ground position , it is computed from the ground speed and wind speed. it is the vector difference betwen the ground speed and the wind-speed.
As we know, that the air craft usually operates in ground as well as in air. so, the ground speed and air-speed usually change based on the wind component. If we consider that there is no wind component then, ground speed= air-speed. If the wind is blowing in the tail-directionof the aircraft then airspeed is greater than wind speed(air-speed=ground-speed+wind speed). If the wind is blowing in the Head direction of the air-craft then airspeed is less than ground speed. So, airspeed= groundspeed-wind speed. gravity also plays a major role in increasing the aircrafts's speed while its desending. If the measured Airspeed is greater than observed ground speed then the wind speed becomes positive. here wind speed is the vector difference between ground and Air-speed. If in case the wind is zero it is equall to ground-speed.
STUDY 3:> Why is it not recommended to use aircraft engine power to move it on the ground at Airport?
The Engine of the aircraft can easily do the work under its ow power to go towards runaways and after landing taxi to their assigned gate. But, as we know that the engine is so powerfull that the thrust developed produces jet blast which might cause damage to the terminal building or various people and equipment. Engines which are close to the ground may also blow sand and debris back and forth which may get sucked to the engine. A pushback is therefore the preferred method to move the air-craft away from the gate and most importantly it reduces the cost of aviation fuel. Taxing can save a lot of money.
- There is a risk of injury to ground support personnel who may be in relatively close to the air-craft.
- It is easier for the truck driver to see the surroundings better than the pilot who is sitting in the cock-pit wit limiting window. the pushback driver knows the airport better than the pilot.
- Effectively a loss of control can happen, especially during high power engine running. Damage can occur to the air-craft itself,other aircraft nearby and other buildings.
- A new system was proposed where a new system in which a push-back truck will move the air-craft all the way to run-way to save fuel and reduce impact of environment.
- The plane engines are very loud, when more than one plane will addup their sound, then it will be unbearable for passengers and other parts near airport.
STUDY 4:> HOW AN AIRCRAFT IS PUSHED TO RUNWAY WHEN ITS READY TO TAKE-OFF ?

TAXIING :
It is defined as the movement of the air-craft on the ground, under its own power where as towing or pushback is where the air-craft is moved by the tug. The air-craft usually moves on wheels but the term also used for the air-craft that has skies or floating equipments. An airplane uses taxiways to taxi from one-place on the air-port to another, example when the air-craft moves from hanger to the runaway.
TOWING :
STUDY 5:>
TAKE-OFF POWER:
In the case of light aircraft full power is used during take-off and in large aircrafts like airliner use a reduced level power to take-off,where less than full power is used to prolong the engine durability. before take-off the piston engines are run routinely at high power to check engine related problems.The nose is raised to a nominal 5-15 degree nose up pitch alltitude toincrease lift from the wings and effect lift off. The takeoff power is controlled in such a way that it can cover 35 feet screen height plays in the take-off. The pilot needs to look around different distances that affect the take-off performance which are:
- TAKE-OFF RUN AVAILABLE(TORA): The distance from which the air-craft can start its take-off run to the point till the end marking for surface change.
- THE CLEARWAY: This is the area st the end of TORA that is free from buildings trees or any other obstructions .The aircraft can use this area to acheive 35 feet sceen height.
- ACCELERATE-STOP DISTANCE AVAILABLE(ASDA): The final distance that must be considered for take-off performance is the ASDA, This distance of weight bearing surface is available to the air-craft to accelerate and then come to stop in case of emergency.
- Take-off power= Force*Displacement i.e (m*dv/dt)*2*S1=> (T-D-F)*S1, where S is the runway length. The forces involved are T= thrust of propulsion system, D= Aerodynamic drag which the vehicle resists, F= rolling resistance due to the contavt on wheels on the ground
TIRE DESIGN:
The tire design of an aircraft may seem a challenging job, because the tyre has to withstand heavy loads while landing, take-off, taxiing and parking. The number of wheels required for an aircraft increases as the weight of the aircraft increases cause it should be distributed evenly. The tyre generates a lot of heat during landing because it drags the wheel to match its momentum. The tyre is designed in such away that it withstands hugh crosswind conditions, extreme pressure and to prevent hydroplanning and for increased braking effect.
There are two types of construction of tyres which are :
Radial tyre: It has fiber strands of the poly fabric at 90degree to the direction of rotation and the tyre side-wall. This restricts flexibility of the bi-deirectional sidewall while it strengthnes the tyre to carry heavy loads . Due to their unique construction radial tyres are resistant to penetration and cuts. Due to stiffer threads they distribute weight evenly. Thus provide greater traction and helps in preventing heat.
Bias tyre: It has the fabric bias oriented with and cross the direction of rotation and the side-wall. Since the fabric stretches, the tire is flexible and can absorb loads, the cut resistnt area is the sidewalls, strength is obtained by adding piles, these tyres doesn't absorb heat well.

Rolling resistance:
It is defined as the force resisting the motion when a body rolls on a surface (ex- ball, tyre) .Rolling resistance can be seen in an aircraft when the air-craft takes off from the runway and while landing on the runway.
rolling resistance force can be written as: Frr=Urr*m*g(N)
Where Urr= coefficient of rolling resistant
m= Mass(kg)
Frr= rolling resistance (N)
g= acceleration due to gravity(9.81m/s^2)
Tyre pressure:
The tyres of an air-craft have to withstand very high amounts of pressure, while take-off, landing, towing, taxing. as the weight of the aircraft increases the no. of wheels are also increased.as we know that while landing the aircraft drags the wheel till it matches with its rotational velocity, due to this a lot of heat is generated due to friction. aircraft tires are usually filled with nitrogen to minimize expansion and contraction from changes in tempretures and pressure during flight , Also the use of inert gasses for tire inflation reduces the possibility of tire explosion. services should be done regularly by checking tyre pressures and maintaining proper tire inflation pressure.
Rolling resistance and tire pressure are inversely proportional to each other.
Brake devices used during landing:
- Wheel brakes(Anti-skid and Auto-brakes): Aircraft have highly-engineered, multi-disc brakes made of a carbon-steel material. Manufacturers include Honeywell and Goodrich, part of Collins Aerospace. In general, the brakes are composed of discs that rotate with the wheel and separate discs that are fixed. When pressure is applied from the right fixed disc on all of the discsthe entire assembly compresses to create the braking action. Carbon brakes are standard today; they cool faster than steel brakes (their predecessor) and can weigh up to 700 pounds less. For the initial part of the braking, say down to around 70 knots, we let the autobrakes do the job, but after that we have to come onto the toe brakes to disengage the autobrake The transition from autobrake to manual brakes can be quite difficult to do smoothly especially in a crosswind, as you will be steering the aircraft with the rudder pedals and then have to press down on the tops of the rudder pedals evenly to deactivate the autobrakes and start manual braking.

- Spoilers: when the wheels touch the ground, a set of spoilers raise up quickly, which kills the lift provided by the wings. (There are two sets seen in the image below: ground spoilers that only deploy on landing, and spoilers that are used in flight as speed brakes as well as on landing.) While the spoilers provide some drag, their primary purpose is to stop the plane from actually flying, putting all of the weight on the landing gear and in turn making the brakes that much more effective.

- Reverse thrust: he final element of braking action is the thrust reversers. In a modern high-bypass turbofan engine, a full 90 percent of the air passing through the engine is simply sucked in, sped up and pushed out of the engine without mixing the fuel. In these engines, ducts on the side of the engine open up and direct that air out and nearly forward, opposite the direction of travel. This is called a cold-stream reverse thrust, and is part of the engine design of most modern engines. (You may have also seen clam-shell style reverse thrusters on older aircraft, which cup the back of the engine.)

STUDY 6:> A. With necessary assumptions, calculate the force and power required to push / pull an aircraft by a towing vehicle
B. Develop the model for the calculated force and power using Simulink.
Solution: Considering the parameters for airbus a380, assumed towing vehicle weight 1000kg


Here, the minimum power required for the towing vehicle to push/pull an aircraft of the given specifications is 45.60kw .So we took the above assumed parameters and developed a simulink model.

- Here we see that the output in the display is 45.653kw.
- Constant blocks are used to feed the parameters such as mass, coefficients and others
- Using sum block total resistant forces are calculated
- Display is used for visuals
STUDY 7:>
- Design an electric powertrain with type of motor, it’s power rating, and energy requirement to fulfill aircraft towing application in Simulink. 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 powertrain.
- Design the parameters in excel sheet.
Solution:
SIMULINK MODEL OF POWERTRAIN OF AN AIR-CRAFT TOWING VEHICLE:

Estimation of Duty cycle: As we know that the power required for towing vehicle to push/pull is equall to 45.60kW. For this model we have considered the motor REB 90 for drive train.

Considering 60kW as continous power, so the voltage will be 400V. Battery power is considered to be 550V. we know that duty cycle is given by d=Vo/Vb. d=400/550, => 0.7272. There will also be losses due to the resistance of FET and inductance. Correction factor is considered to be 90% ,that makes us consider that there is 10% loss, So the voltage which is demanded by the motor should be 10% more than the required voltage.[d =0.7272*(1+0.1)= 0.799~0.8. so here the voltage demand is 440V from the battery.
| Mass of the aircraft | 5,80,000kg |
| Mass of towing vehicle | 1000kg |
| Coefficient of rolling resistance | 0.002 |
| Drag coefficient | 0.025 |
| Air-density | 1.125kg/m^3 |
| Frontal area | 150m^2 |
| Velocity | 4m/s |
| Uphill gradient | 0 deg |
| Battery voltage | 550V |
| Correction factor | 0.8 |
INPUT:

OUTPUT FROM THE BATTERY :

Output from the motor:

After simulating the model the parameters are stated in excel sheet:

Google drive link for my models and excel sheet: https://drive.google.com/drive/folders/1jMHkgEDvTXdft2yKcQA-FhOl3CnygzF4?usp=sharing
References:
https://en.wikipedia.org/wiki/Aircraft grossweight
https://www.grc.nasa.gov/www/k-!2/airplane.html