Lets go to ATHS

airport

G12 ES Term IDifferentiation and its applications

Group Members

• Sulaiman Murad

• Hamdan Khaled

• Abdulla Mousa

THE HISTORY OF AIRBUSA380

• The 555 seat, double deck Airbus A380 is the world's largest airliner, easily eclipsing Boeing's 747. The A380 base model is the 555 seat A380-800 (launch customer Emirates). Potential future models include the 590 ton MTOW 10,410km (5620nm) A380-800F freighter, able to carry a 150 tonne payload, and the stretched, 656 seat, A380-900.

• Airbus first began studies on a very large 500 seat airliner in the early 1990s. The European manufacturer saw developing a competitor and successor to the Boeing 747 as a strategic play to end Boeing's dominance of the very large airliner market and round out Airbus' product line-up.

• Airbus began engineering development work on such an aircraft, then designated the A3XX, in June 1994.

Airbus A380 characteristics.

http://www.slideshare.net/rubal_9/airbus-a380-1856096

How long does it take to take off, and the velocity of it

http://www.youtube.com/watch?v=p4DTormtEG0

• Generally, 8000 feet is long enough for anything that flies nowadays. the a380 was designed, as much as possible, to be compatible with 747 infrastructure.

What is the take o f f distance , i . e. t h e runway length that willbe crossed before it takes off ?

Speed of an airplane by Radar

Speed of an airplane by Radar

• X2+Y2=52

X2+62=52

X2+62=(10)2

X2=102-62=64X=√64 =8X2+Y2=52

2X×(dx/dt)+2y×(dy/dt)=2s×(ds/dt) *2y×(dy/dt)=0 because the height don’t change2(8)×(dx/dt)=2(10)×(400)(dx/dt)=8000/16=500

Waiting at the airport

Waiting at the airport

• ω(x)=X2/2(1-X)ω|(x)=(2-2x)(2x)-X2(2)×(-1)/4(1-x)2

ω|(x)=-4x2+4x+2x2/4(1-x)2

ω|(x)=-2x2+4x/4(1-x)2

=8x(2-x)/4(1-x)2

=x(2-x)/2(1-x)2

ω|(0.1)=(0.1)×(2-0.1)/2(1-0.1)2

=0.19/1.62 =0.1173ω|(0.7)=(0.7)×(2-0.7)/2(1-0.7)2

=0.91/0.18 =5.0556•

Taking off Distance and Time

Taking off Distance and Time

• D(t)=(10/9)t2

D|(t)=20/9 ×t=72t=72/(20/9)=32.4sD(32.4)=(10/9)(32.4)2=1166.4m

Aircraft Glide Path

A.G.P• When x=-4 , y=1• F(x)=9x3+bx2+cx+d

• Also, when x=0, y=0 when hitting the origin, 50 , d=0• 50, f(X)=9x3+bx2+cx+0• 1=-64a+16b-4c• -1=64a-16b+4c

• f’(x)=3ax2+26x+c ; [-4,-1]• f’(-4)=0 , f’(0)=0• When x=0 then 3(0)+2(0)+c=0c=0

Part a

A.G.P

• When x=-4, 48a-8b=048a=8b6a=b• From before , 64a-16b+4(0)=-164a-16(6a)=-164a-96a=-1-32a=-1a=1/32• b=6a=6x1/32=3/16

Part a

F(x)=1/32x3+3/16x2

A.G.P• f’(x)=3/32x2+3/8xf’’(x)=3/16x+3/8=03/16x=-3/8X=-2• When x=-2f(-2)=3/32(-2)2+3/8(-2)=(3/8)-(3/4)=-3/8• When x=-2f(-2)=1/32(-2)3+3/16(-2)2

=-1/4+3/4=1/2

Part b

So the max rate would be at (-2,1/2)

Shock wave

• F is frequency perceived by the observer.• fs is the frequency of the source.• Vw is the velocity of the waves.• VL is the velocity of the listener.• Vs the velocity of the source.

• A) Find an equation of the F in term of Vs.• I Will assume that this is a A380 Airbus.• V = speed of waves (sound) = 340 m/s• VL = 0, since he isn’t moving.

• Where L is the length of the airbus and g is the gravity of earth

• Therefore,

• Which is :

• b)

• c)

Finally we arrived