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1. There are two organisations which that deals with the categorization of aircrafts based on their weights namely, FAA – Federal Aviation Administration ICAO – International Civil Aviation Organization (HQ: Montreal, Canada) While FAA is U.S.A specific, ICAO governs the entire world in the aircraft domain.…
Sagar M Iyengar
updated on 29 Oct 2020
1. There are two organisations which that deals with the categorization of aircrafts based on their weights namely,
FAA – Federal Aviation Administration
ICAO – International Civil Aviation Organization (HQ: Montreal, Canada)
While FAA is U.S.A specific, ICAO governs the entire world in the aircraft domain. Other than that, their rules and policies are very much similar.
So, let’s take a look at the ICAO categorization’s categorization of aircrafts based on their weights:
According to ICAO, there are 3 categories of aircrafts based on their weights:
Abbreviations: WTC – Wake Turbulence Categories, MTOW – Max Take-Off Weight, MLW – Max Landing Weight, TOR – Take-Off Run, LR- Landing Run
Name MTOW (Kgs) MLW (Kgs) TOR(m) LR(m)
(Heaviest aircraft to be recorded 640000 591700 3500
A cargo aircraft designed by Antonov
Design Bureau (ADB) )
Stratolaunch (aircraft features twin fuselage
And longest wingspan and is used for
Air air-launch-to-orbit rockets.)
(Largest passenger airliner)
(Latest and largest version of 747)
-Super Sonic Transports WITH
a Mach of 2.15 and a passenger
capacity of 150)
SSTs)
(Aerospatiale/BAC
British-French turbojet powered
Passenger supersonic airline with a Mach = 2.04
With passenger capacity of 98-128)
Name MTOW (Kgs) MLW (Kgs) TOR(m) LR(m)
(narrow body airliner built by American
Douglas company)
(narrow body jet airliner by Convair
Designed to compete with Boeing 707.
It was once a fastest jet with a speed of 990 kmph)
Airbus A320 family intend for short range services)
Speed turbo-prop airliner which cruise at a speed of 665 kmph)
Name MTOW (Kgs) MLW (Kgs) TOR(m) LR(m)
EMB-500 (light jet developed by Brazilian aircraft manufacturer Embraer)
2.
Yes, there is a sizeable difference between Air speed and ground speed as elucidated below:
Ground speed – It is the speed of aircraft relative to earth’s surface
As this speed is relative to the ground, its is considered as horizontal speed of the aircraft rather than vertical speed. Technically speaking, all the lift components are neglected and only the drag component is taken into consideration. So, an aircraft climbing completely vertical will have zero ground speed
Ground speed is what determines how fast an aircraft will reach its destination.
Air speed (a.k.a TA-True Airspeed) – Speed of the aircraft relative to air it is flying in. We can
also say that it is also the speed of the air which is flowing around the aircraft
It is a parameter that determines whether there is enough air flow around the aircraft top make it fly
Relation between the two:
Ground speed = Air speed ± Wind speed
Ground speed = Airspeed + windspeed
Ground speed = Airspeed - windspeed
Air speed is always 25% higher than Ground speed.
3.
There are three main reasons to avoid using engine propulsion power to move it on the ground:
The amount of power available from aircraft engine propulsion is way too high compared to the power required for taxiing or the power required to move it on the ramp area.
By using the aircraft power on ground, a great deal of damage will be dealt to the engine components because the piston and other components will be fatigued every time the engine is braked. This is mainly because, as the engine runs, it reaches very high power within no time, and very soon it has to be braked to keep it from colliding into something. So every time the engine is braked from such a massive power, the engine components receive a massive blow.
In tight and crowded areas, blasting jet reversers can fan out huge amount of debris into the crowd or into other vehicles and aircrafts which can cause what is known as Foreign Object Damage (FOD).
Moreover, the exhaust from the aircrafts even at subsonic speeds will be around 166 m/s which is almost half the velocity of sound at 0 degree Celsius and is enough to deafen someone within a 500 meters radius. Imagine propelling a Supersonic Transports like Concorde or a TU-144. They can possibly kill a handful of people with the supersonic blast emerging from their exhaust.
For the above reasons, aircrafts are generally forbade from using their engine power on the ground.
4. In aircraft terminology, pushback is something that happens when an aircraft is ready to take-off and is pushed backwards away from the airport gate by vehicles called tugs or tractors.
At the moment of departure, an aircraft tug will park in front of the aircraft nose wheel to start pushing it to the runway.
Either the tug is directly attached to the plane’s nose gear with a tow bar or could be a “wheel-lift” tug. These tugs cradle the nose gear and lift it up before moving the plane which yields the tug driver control over plane’s direction during pushback.
New taxi technologies are coming into existence like pilot-controlled tugs and electric-motors mounted to the plane’s landing gear which has the advantages of high fuel efficiency and reduction in noise.
5.
1. Take-Off power:
Take-off can be defined as the action of getting an aircraft airborne and the power required to achieve this is called take-off power.
For flights with horizontal take-off, they should run for a certain distance (usually the runway spans from 2000-3500 meters) before getting airborne. Take-off power can be defined as a function of Maximum Take-Off Weight (MTOW), density of the ambient air and the available power from the propulsion unit.
Take-Off power is almost never maximum engine power as it affects engine life and brings substantial wear to engine components.
2. Tyre design:
Thumb rule for aircraft’s tyre requirement is that, heavier the aircrafts, more will be the number of tyres in it in order to even out the aircraft weight uniformly on the ground and maintain stability.
The tread pattern of aircrafts is made in such a way that they channelise the water flow smoothly and avoid hydroplaning (uncontrolled sliding of vehicle due to tires encountering more water than the treads can displace. This happens if the treads are worn out) and render stability during crosswinds.
Casing plies run diagonally at at approximately right angles to each other.
Cross-ply tyres consist of carcass layers made from nylon cord. They are placed diagonally across each other in the tread and the sidewalls, at an angle of 55 degrees. Multiple rubber plies overlap each other and they form a thick layer, resulting in less flexibility which can make it more sensitive to overheating.
Aircraft tyres are generally Inter Tread Reinforced fabric (ITF tyres). This provides the tyres stability at high speeds and lesser tread distortions under load.
In radial tyres the casing plies run radially from bead to bead at approximately 90 degrees to centre-line of the tyre.
Radial tyres were developed in 1946 by Michelin. At the time there was a need for more flexible tyres which were able to absorb shocks generated by road surfaces. The sidewall of radial tyres and the tyre tread work as two independent features. The flexibility of a Radial tyre, together with its strength, are two combined factors which mean a radial tractor tyre absorbs impact shock and bumps more effectively than a cross-ply tyre.
3. Rolling resistance:
Rolling resistance experienced by any vehicle’s tyre is nothing but the resistance offered by the surface on which it rolls to the vehicle’s tractive power.
Formula:
Frr = murr * m * g
Where,
murr is the rolling resistance coefficient and m and g are mass and acceleration due to gravity.
For an aircraft, it is influenced by many factors like tyre pressure, tyre temperature, landing speed, Maximum landing weight (MLW) Runway roughness etc.,
4. Tyre pressure:
Aircraft tires generally operate at high pressures, up to 200 psi for airliners, and even higher for business jets. The main landing gear on the Concorde was typically inflated to 232 psi, whilst its tail bumper gear tires were as high as 294 psi.
5. Brake force while landing:
The braking of an aircraft while landing can happen in three different ways and are explained below:
In the first method: the landing gears are braked with help of disc rotor brakes which are generally hydraulic in nature.
Second method: A reverse thrust is applied to slow down the aircraft. Basically, direction of propulsion is reversed here.
Third method: A special type of brakes called air brakes is used to stall the aircraft by increasing the drag force that is acting on the frontal area of the aircraft.
6. The power required to push or pull the aircraft by a towing vehicle can be computed as follows:
Let’s take the example of concorde for the calculation purpose.
M = 185000 kgs
Take g to be 9.81 m/s^2
For aircraft tug, consider AL-250.
M = 8981 kgs
First, the tractive effort required to pull the aircraft by the tug should be calculated:
Frr = murr * m * g (as explained earlier)
Assume,
murr for concorde = 0.008
murr for AL 250 = 0.002
rolling resistance of both aircraft and tug:
Frr = (0.008*185000 + 0.002*8981) * 9.81
Frr = 14.695 kN
2. Air drag force:
Density of air at 20 degree Celsius, rho = 1.204 kg/m3
AL-250 velocity, V = 6.44 KPH = 1.78 m/s (Loaded)
Concorde Frontal area, A = 32 m^2
Drag coefficient Cd = 0.018
Fd = 0.5*Cd*rho*A*V2 = 61.03 N = 0.061 KN
Ergo, Tractive effort require by the tug is,
Fte = Frr + Fd = 14.695 + 0.061 = 14.756 KN
Power required:
P = Fte * V = 14.756 * 6.44
P = 95.02 KW.
7. From the above results, we got to know that the power requirement for the tug is 95.02 KW
Let’s consider Marlin 13 13KW motor for each wheel which costs around 6660 $ = 466200 Rs. Since there’s one for each wheel, it adds up to around 1864800 Rs.
A table of assumed parameters is given below:
Assumption parameter |
Value |
Aircraft tug cycle time |
20 mins = 0.33 hours |
Battery – LiNiMnCoO2 |
120 KWh |
4 Marlin 13 72v 13 Kw Lynch motor |
4*72 = 288v |
Peak power rating |
26*4 = 104 Kw |
Current rating |
200 amp |
Rated speed |
3240 rpm |
Peak torque |
72 N-m |
Power converter |
DC-DC chopper |
Gear ratio, G |
22:56 |
Tyre radius |
18 inches = 0.4572m |
Torque requirement:
Fte = T*G/r
Fte = 14.756 KN from previous calculation
Therefore, T = 14.756 * 22 / 0.4572 = 710.04 N-m
Let’s use Lithium Nickel Manganese Cobalt Oxide battery which punches a power of 120 KWh.
Input power required by the motor:
Pi = 72 * 200 = 14400 = 14.4 KW
For 4 motors,
Pi = 4 * 14.4 = 57.6 KW
Hence battery satisfies the power requirement of the 4 motors with a buck convertyer in between
Input power rating for assumed cycle time of 0.33 hrs,
Pin = 57.6 * 0.33 = 19 KWh
Output power rating for assumed cycle time of 0.33 hrs,
Po = 95.02 * 0.33 = 31.35 KWh
Total power rating, Pt = Po + Pin = 19 + 31.35 = 50.35 KWh
Therefore,
Duty cycle,
D = Po/Pt = 31.35 / 50.35
D = 0.6226 = 62.26 %
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