Project-1: Powertrain for aircraft pushback trucks in runways

Question 1: List out the total weight of various types of aircrafts.

Airplanes

MTOW = Maximum take-off weight, MLW = Maximum landing weight

Type	MTOW [kg]	MLW [tons]
Antonov An-225	640,000	591.7
Scaled Composites Model 351 Stratolaunch	589,670
Airbus A380-800[1][2]	575,000	394
Boeing 747-8F	447,700	346.091
Boeing 747-8	443,613	306.175
Boeing 747-400ER	412,770	295.742
Antonov An-124-100M	405,060	330
Boeing 747-400	396,900	295.742
Lockheed C-5 Galaxy[3][4][5]	381,000	288.417
Boeing 747-200[6]	377,840	285.700
Boeing 747-300[6]	377,840	260.320
Airbus A340-500[7]	371,950	240
Airbus A340-600[7]	367,400	256
Boeing 777F	347,800	260.816
Boeing 777-300ER	351,800	251.29
Boeing 777-200LR	347,450	223.168
Boeing 747-100[6]	340,200	265.300
Airbus A350-1000	308,000	233.5
Boeing 777-300	299,370	237.683
Boeing 777-200ER	297,550	213.00
Airbus A340-300[7][8]	276,700	190
McDonnell Douglas MD-11	273,300	185
Airbus A350-900	270,000	175
Ilyushin Il-96M	270,000	195.04
McDonnell Douglas DC-10	256,280	183
Boeing 787-9[9]	254,000	192.777
Boeing 787-10[9]	254,000	201.849
Airbus A340-200[7][10]	253,500	181
Ilyushin IL-96-300	250,000	175
Airbus A330-300[11][12]	242,000	185
Airbus A330-200[11][12]	242,000	180
Lockheed L-1011-500	231,300	166.92
Boeing 787-8[9]	228,000	172.365
Lockheed L-1011-200	211,400
Ilyushin IL-86	208,000	175
Boeing 767-400ER	204,000	158.758
Airbus A300-600R[13]	192,000	140
Boeing 767-300ER	187,000	136.08
Concorde	185,000	111.1
Airbus A300-600[13]	163,000	138
Boeing 767-300	159,000	136.078
Airbus A310-300[13]	157,000	124
Vickers VC10	152,000	151.9
Boeing 707-320B[14]	151,000	97.5
Boeing 707-320C[14]	151,000	112.1
Douglas DC-8-61	147,000
Airbus A310-200[13]	142,000	123
Airbus A400M	141,000	122
Douglas DC-8-32	140,000
Douglas DC-8-51	125,000
Boeing 757-300	124,000	101.6
Boeing 707-120B[14]	117,000	86.3
Boeing 757-200	116,000	89.9
Boeing 720B[15]	106,000	79.5
Boeing 720[15]	104,000	79.5
Tupolev Tu-154M	104,000	80
Tupolev Tu-204SM	104,000	87.5
Convair 880	87,500
Boeing 737-900	85,000	66.36
Boeing 737-900ER	85,000	71.35
Boeing 727-200 Advanced[16]	84,000	70.1
Airbus A321-100[17]	83,000	77.8
Boeing 737-800	79,000	65.32
Boeing 727-200[16]	78,000	68.1
McDonnell-Douglas MD-83	73,000	63.28
Boeing 727-100[16]	72,500	62.4
Boeing 727-100C[16]	72,500	62.4
McDonnell-Douglas MD-90-30	71,000	64.41
de Havilland Comet 4	70,700
Boeing 737-700	70,000	58.06
Airbus A320-100[17]	68,000	66
Boeing 737-400	68,000	54.9
de Havilland Comet 3	68,000
Boeing 377	67,000
Boeing 737-600	66,000	54.66
Airbus A220-300	65,000	57.61
Hawker Siddeley Trident 2E	65,000
Airbus A319[17]	64,000	62.5
Boeing 737-300	63,000	51.7
Boeing 737-500	60,000	49.9
Airbus A220-100	59,000	50.80
Airbus A318[17]	59,000	57.5
Boeing 717-200HGW	55,000	47.174
Douglas DC-7	55,000
de Havilland Comet 2	54,000
Boeing 717-200BGW	50,000	46.265
de Havilland Comet 1	50,000
Douglas DC-6A	48,600
Douglas DC-6B	48,500
Embraer 190[18]	48,000	43
Caravelle III	46,000
Fokker 100	46,000	39.95
Douglas DC-6	44,000
Avro RJ-85	42,000	36.74
Handley Page Hermes	39,000
Embraer 175[19]	37,500	32.8
Bombardier CRJ900[20]	36,500	33.345
Embraer 170[21]	36,000	32.8
Bombardier CRJ700	33,000	30.39
Douglas DC-4	33,000
Vickers Viscount 800	30,400
Bombardier Q400	28,000	28.01
Bombardier CRJ200	23,000	21.319
ATR 72-600	22,800	22.35
Saab 2000	22,800	21.5
Embraer ERJ 145	22,000	19.3
ATR 42-500	18,600	18.3
Saab 340	13,150	12.930
Embraer 120 Brasilia	11,500	11.25
BAe Jetstream 41	10,890	10.570
Learjet 75[22]	9,752	8.709
Pilatus PC-24[23]	8,300	7.665
Embraer Phenom 300[24]	8,150	7.65
Beechcraft 1900D	7,765	7.605
Cessna Citation CJ4[25]	7,761	7.103
de Havilland Hercules	7,000
Embraer Phenom 100	4,800	4.43

maximum takeoff weight :

The maximum takeoff mass (MTOM), often referred to as maximum takeoff weight (MTOW), of an aircraft is a value defined by the aircraft manufacturer. It is the maximum mass at which the aircraft is certified for take off due to structural or other limits. MTOW is usually specified in units of kilograms or pounds.

maximum landing weight :

The maximum landing weight (MLW) is the maximum aircraft gross weight due to design or operational limitations at which an aircraft is permitted to land.

Question2: Is there any difference between ground speed and air speed?

yes

Ground speed :

Ground speed is the horizontal speed of an aircraft relative to the ground.

An aircraft heading vertically would have a ground speed of zero.

Air speed:

Airspeed is the speed of an aircraft relative to the air.

True airspeed :

True airspeed is the airspeed of an aircraft relative to undisturbed air.

Lets understand these terms with an example.

Consider the above figure

In the above figure the speed of the aircraft is 50 knots in the still air condition.

Therefore the speed relative to the ground will be 50 knots.

And the speed relative to the air will be 50 knots.

In the above figure the speed of the aircraft is 50 knots.

And a tail wind is blowing at a speed of 10 knots.

The wind will increase the speed of the aircraft relative to the ground to 60 knots.

In the above figure the speed of the aircraft is 50 knots.

And a head wind is blowing at a speed of 10 knots.

The head wind will decrease the speed of the aircraft relative to the ground to 40 knots.

Therefore, from the above examples we can have understanding what airspeeds and ground speeds are.

Question 3: Why is it not recommended to use aircraft engine power to move it on the ground at Airport?

There are number of reasons to not use engine power on the ground.

1. Jet blasts due to engines revving at high rpms will cause damage and uncertainity to the surrounding and personnel.

2. In low engine mounted airctrafts the engines will start picking up the foreign particles along with the air like gravels,snow,grass etc

this will cause engine damage.

3. in high mounted engines there is a issue with the visibality of the rear view due to the engine placement which affects the reverse movement of the aircraft.

4. usually to move the planes in reverse direction the engines need to rev in higher rpms this will require more fuel consumption.

when oil prices are low it is manageable but in other situations it’s not an economical solution.

5. due to high revving nature of the engines during reversing the noise it creates cannot be tolerated by the people in the surroundings.

this will create an unpleasant environment.

6.engine wear

Plane engines are designed to run at pretty constant power for long periods.

Power back requires a short burst of high thrust before going back to idle.

The fact this is done when the engine has not long been started and might not be up to temperature fully will cause engine wear in the long term if the engine is used for reversing regularly.

Question 4: How an aircraft is pushed to runway when its ready to take off?

due to the above-mentioned reasons an aircraft engine cannot be used for manoeuvring the aircraft in an airport full of personnel and passengers.

Therefore, a push back truck is used to perform the operation of pushing the aircraft to the runway.

there are different types of pushback trucks-

1. truck which uses a tow bar and its own weight for the necessary traction to perform the operation.

2. trucks which doesnt use a towbar and uses the weight of the aircraft to perform the operation.

The below images are

Douglas TBL – 400 pushback towbar-less tug.

The towbar less tugs use the weight of the aircraft to gain necessary traction to perform the push back operation.

There is a hydraulic system which holds the nose wheel of the aircraft and uses the nose wheel weight for the necessary downforce which will provide enough traction to perform the pushback operation.

please refer the below link to obsever the operation of a towbar less pushback truck performing pushback operation.

https://www.youtube.com/watch?v=vR97Vj4iBJg

The hydraulic system will hold the nose wheel of the aircraft and the driver of the truck will be in control during the pushback operation.
The driver will carefully observe the surroundings and with proper communication with the air traffic control they perform the pushback operation.

5.Learn about take off power, tyre design, rolling resistance, tyre pressure, brake forces when landing.

take off power:

The power required by the engine to achieve a certain speed of the aircraft so that the air passing over the wings will generate enough lift to lift the weight of the aircraft and take it to atmosphere is called a take off power.

The 100 % thrust is not used while performing this operation because of the following reasons-

The turbo fan jet engines mounted on the either side of the aircraft have a tendency to accelerate at slightly different rates up to 40 % thrust .

That means to reach the 40% thrust there will be a small time difference between the engines if we apply the full thrust immediately this time lag will continue up to 90% of the thrust and makes the plane really unstable.

So the engines are made to 40% thrust and after that the turbo fan jet engines are almost identical then the thrust is increased to take off power.

This process is called stabilizing of the engines.

This process is used by large capacity aircrafts.

tyre design:

An aircraft must withstand a wide range of operational conditions.

When on the ground, it must support the weight of the aircraft. During taxi, it must provide a stable, cushioned ride while resisting heat generation, abrasion and wear. At take-off, the tire structure must be able to endure not only the aircraft load but also the forces generated at high angular velocities.

Landing requires the tire to absorb impact shocks while also transmitting high dynamic braking loads to the ground. All of this must be accomplished while providing a long, dependable, reliable, service life.

Tire prototypes are put through rigorous tests that simulate cycles of landings, take-offs, and taxi operations. It is only after the tires pass all mandatory tests and meet the requirements of the aircraft manufacturer and airworthiness authorities that the tires are put into production.

tires are designed as a composite of various rubber, fabric and steel products.

There are 2 types of tyres used-

1. cross‑ply or BIAS tire
2. RADIAL tire

Nomenclature of the tyre.

The “chine” tire is a nose wheel tire designed to deflect water and slush to the side and away from intakes on aft­-fuselage mounted jet engines.

It consists of a flared upper sidewall protrusion which deflects the spray pattern of water or slush displaced by the tire's contact with the runway. A tire can have a single chine (one sidewall flared) for dual nose wheel tire configurations or double chines (both sidewalls flared) for single nose wheel tire configurations.

Rolling resistance:

Rolling resistance, sometimes called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface.

It is mainly due to the visco elastic nature of the tyre

When the tyre is rolling its threads goes through a series of loading and unloading.

While loading and unloading if we observe the stress strain curve, they both are different.

I.e. the thread will not follow the same path of loading curve while unloading.

We can understand the phenomenon from the image.

In the image, we can observe a hysteresis loss due to different paths of loading and unloading of the thread of a tire.

I.e. the threads will take more time to regain their original size and shape due to the visco elastic nature of the rubber.

This is the major reason for the tires to have rolling resistance.

Usually this resistance will consume 30% of the fuel in an automobile.

Tyre pressure

Aircraft tires generally operate at high pressures, up to 200psi for airliners, and even higher for business jets.

Aircraft tires are usually inflated with nitrogen to minimize expansion and contraction from extreme changes in ambient temperature and pressure experienced during flight.

The use of an inert gas for tire inflation will eliminate the possibility of a tire explosion.

The passenger car and truck tires are able to operate continuously because they reach thermal equilibrium while aircraft tires do not.

An aircraft tire which taxis continuously at rated load and 40 mph will continue to heat up until the tire fails.

The failure to reach thermal equilibrium is a consequence of the high deflections at which aircraft tires operate.

High inflation pressure and high deflection rates – 2.5 times greater than a car tire – allow relatively small tires to support the high loads of an aircraft.

It’s important to check tire pressures when they are cold if at all possible.

the below image represents recommended actions when checking air pressures in terms of percent of the operating pressure.

Break force while landing :

The break force is applied in 3 different ways in an aircraft

1.By using the carbon disk breaks which are mounted to the wheel assembly.

Aircraft disc brakes in the landing gear, used to brake the wheels while touching the ground.

These brakes are operated hydraulically or pneumatically.

2.reverse thrust.

The reverse thrust is used in case of slippery surfaces like snow or water.

The turbo jet engines rotate in opposite direction producing a reverse thrust and applying the necessary breaking force required for a safe landing.

3.ground spoilers.

These are the aerodynamic breaks which create drag by the help of the spoilers which helps in breaking for the safe landing.

when the aircraft is approaching the runway, there are aerodynamic forces and moments acting on it as well as the gravitational force.

as soon as the aircraft wheels touch the runway, there appears a new force acting on the aircraft while other forces are changing.

The wheel force of friction starts to act on the aircraft.

The aerodynamic forces start to decrease.

The wheel friction force depends on the magnitude of aerodynamic lift.

6.With necessary assumptions, calculate the force and power required to push / pull an aircraft by a towing vehicle.

While calculating the forces acting on a vehicle we consider the following forces-

1.rolling resistance force

2.aerodynamic drag force

3.hill climbing force

4.acceleration force

But in the process of push back the speed in which the operation is performed is very less its around 10 kmph so at that speeds the aerodynamic forces acting on the vehicles is negligible.

And as the airport runways are built with zero grade angle with extreme caution due to safety reasons the hill climbing force will be zero and can also be neglected.

And the push back operation is a uniform speed operation the acceleration force acting on the vehicle can be neglected.

Finally, the rolling resistance force should be considered while calculating the total force.

Considering the Boeing 737-900 aircraft and Douglas TBL – 400 push back truck we shall find the force required for the pushback operation.

Mass of the Boeing 737-900(MTOW) = 85000 kg

Mass of the Douglas TBL-400 =29,900kg

Weight =(mass of aircraft + mass of the pushback truck)*9.81

=(85000+29900)*9.81

=1127169 N

Assuming the rolling resistance coefficient = 0.06

Force=weight*rolling resistance

=1127169*0.06

=67630.14 N

FORCE=67630.14 N

Assuming the maximum speed the truck performs the push back operation = 10kmph=2.77m/s

Power = force * velocity

= 67630.14*2.77

=187861.5 watts

POWER=187861.5*0.001341 HP

=251.92 HP

POWER = 187861.5 Watts = 251.92 HP = 187.8615 KW

assuming the wheel diameter of the pushback as 5 feet (1.524 m) radius = 0.76 m

torque = force * radius

= 67630.14 * 0.76

Torque = 51534.16 Nm

vedio refered to understand about the pushback operation : https://www.youtube.com/watch?v=7Rmi30A7TFw&t=217s

7. Design an electric powertrain with type of motor, it’s power rating, and energy requirement 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 powertrain.

assuming a single pushback operation takes place for 20 min (0.33 hours).

Energy = power (kw) * time ( hours)

=187.8615 * 0.33

= 61.99 kwh

energy to perform a single pushback operation = 61.99 kwh

Assuming the pushback truck works for 8 hours per day we can estimate the energy capacity which it requires .

Energy = power (kw) * time (hours)

= 187.8615 * 8

= 1502.892 Kwh

ENERGY = 1502.892 Kwh

from the above calculations we use a brushless dc motor that should have a rated power of 250 HP and should be capable of producing 51534.16 Nm of torque

and the battery capacity should be 1502.892 kwh to perform a single day push back operation.

By observing these numbers we can try to optimise the pushback truck.

Tesla model s P100 D has a battery capacity of 100kwh lithium ion battery and weighs 625kg.

If we can approximate the mass of the battery used by the truck.

Mass of the battery = (1502.892*625)/100

=9393.075 kg

If we try to replace conventional gasoline powertrain in the Douglas TBL-400 pushback we can reduce the weight of the truck to some extent.

this will reduce many structural elements and reduce the costs.

and compared to gasoline powertrains electric powertrains are more efficient.

This is a economical solution if we can scale the product and reduce the price.

As the majority of weight in an electric powertrain depends on the battery, we can bring down the total mass of the pushback truck from 29900kg to the range of 15000 kg – 20000kg.

To calculate the duty cycle we should consider the rated power to achieve the highest speed.

Considering input voltage of 330 v

Efficiency = Pout/Pin

Pout = Vo * Io

Pin = Vi * Ii

Pout=output power

Pin=input power

Vo=output voltage

Vi=input voltage

Io=output current

Ii=input current

Usually the dc - dc converters used in an electric powertrain are very efficient.

to operate the dc motor at the required power we will use different controllers and converters.

the required voltage values are controlled by different semiconductor swintching circuits.

Let us assume the efficiency of converters and the controlers as 95%

Pin = Vi * Ii

187861.5 = 330 * Ii

Ii=569.277 amps

Input current and output current is considered same

Therefore Io=569.277 amps

Vo=efficiency * Vi

Vo=0.95*330

Vo=313.5 v

Pout=Vo*Io

Pout=313.5*569.277

Pout=178468.33 watts

Duty cycle = Vo/Vi

=313.5/330

=0.95

Therefore the duty cycle range is (0 , 0.95)

list of assumed parameters -

rolling resistance	0.06
max speed of the pushback truck during the operation	10 kmph (2.77m/s)
wheel diameter of the pushback truck	0.76 m
time to perform single pushback operation	20 mins
operating time of the pushback truck in a day	8 hours
input volatge	330 v
efficiency of the converters	95%

BLOCK DIAGRAM :

conclusion :

the pushback trucks gasoline powertrains can be replaced with the electric powertrains and we can increase the efficiency and reduce some structural elements of the pushback truck.

the electric powertrains are known for producing high amounts of torque which is the most crutial aspect for this operation ,there are many products already availabe in todays market which have a electric powertrain.

a company called Mototok have such push back tugs with the electric powertrain and are remote controlled.

if we observe these electric tugs they are really compact yet are capable of performing the operation very effectively.

if we can scale the product we can reduce the overall price and this will be a very economical solution as compared to conventional push back truck.

refer the link to uderstand and observe these tugs performing push back operation.

https://www.youtube.com/watch?v=2wjt8DQQS4k