AS 310: Aircraft Performance

3

Exercise 1: Rectilinear Motion

Uniformly Accelerated Rectilinear Motion and Newtons Law of Momentum

Equations to use (remember to keep track of units):

F = ma

m = W/g

T-D-f = ma

VF 2 = VI 2 + 2 a s VF = VI 2 + a t

Takeoff distance (s) = VF 2 /2aKE = mV2

PE = WhHP= T*Vkts /325

sin() = (ROCkts)/(Vkts)

1 kt = 1.69 ft/sec

g = 32.2 ft/sec2Givens: (Questions 1 to 8)

Gross Weight

= 100,000 pounds

Average Drag

= 5,000 pounds

Average Friction Force = 1,000 pounds

Average Thrust

= 34,000 pounds

Lift Off Speed

= 150 Knots True Airspeed1. Compute the acceleration on the aircraft during the takeoff roll (ft / sec 2). m = W/g

m = 100,000/32.2

m = 3105.6 lb-sec2/ft or SlugsT D f = 3105.6 (a)

34,000 5000 1000 = ma

28000/3105.6 = 9.01 ft / sec 22. What would be the length of the takeoff run (ft)?

Knots to ft/sec = 150x1.69 = 253.5 ft/sec

Takeoff distance (s) = VF2 /2aS = 253.52 /2(9.01)

S = 64262.3/18.02S = 3566.1 ft3. How long would it take until liftoff once the takeoff roll is started (sec)?

V F = VI 2 + a t253.5 = 02+ 9(t)

T = 253.5/9

T = 28.11 sec4. Given the information shown above, determine how fast this airplane should be going when it passes the 2000-foot runway marker (2000 feet from the start of the takeoff roll)? Please express your answer in knots.VF2 = VI2 + 2 a s VF2 = 02 + 2(9.01)(2000)VF2 = 36040

VF2 = 36040

VF = 189.8 ft/ sec

VF = 189.8/1.69 = 112.3 Knots 5. What is the power (HP) of the aircraft engines after takeoff at Average Thrust at 250KTAS?After takeoff, so friction is now gone.

HP= T*Vkts /325HP = 34,000 x 250 / 325

HP = 26,154 HP6. What is the Kinetic Energy (ft-lb) of the aircraft after climbing out at 250 KTAS with the new weight at 95,000 lb?m = W/gm = 95000/32.2

m = 2950.3 Kg

250 Knots = 250 x 1.69 = 422.5 ft/Sec

KE = mV2KE = (2950.3)(422.52)KE = 263 323 494.7 ft-lb7. What is the Potential Energy (ft-lb) of the aircraft after climbing out to 10,000 ft above sea level with the new weight at 95,000 lb?PE = WhPE = 95000 x 10000PE = 950 000 000 ft-lb8. What is the Angle of Climb (deg) for airplane at 250 KTAS with a climb rate of 4,000 ft/min?Sin = O/HSin = 39.4/250 = Sin-1(39.4/250)

= 9.06250 KTAS

4000 fpm

= ?

Change to Knots!!4000 fpm = 66.6 fps = 39.4 Knots

O

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This document was developed for online learning in ASCI 309.

File name: Ex_1_ Rectilinear_Motion

Updated: 05/30/2014

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