TAKEOFF FUNDAMENTALS

Takeoff and Landing are Critical Evolutions

·  Takeoff: about 1/5 (20%) of all accidents

·  Landing: about 1/4 (25%) of all accidents

·  Both evolutions comprise a very small percentage of total flight time

Predicting Takeoff Performance

·  If we know

Takeoff distance X (brake release to liftoff)

o  Take off time t (elapsed time from brake release to liftoff)

o  Takeoff speed V (liftoff speed; no-wind ground speed.

·  Can predict changes in these parameters resulting from changes in

o  Gross Weight

o  Density Altitude

o  Temperature

o  Runway Wind

o  Runway Slope from Horizontal

·  Changes can be very large. For example (a real world example):

o  T: 0o C to 40o C

o  PA: SL t0 2000’ MSL

o  Wind: 25 KT headwind to calm

o  Takeoff Distance: 2500’ to 10,500’ (420% increase)

Deriving Equations for Predicting Changes

Rectilinear motion (t in seconds and X in feet Þ V in ft/sec and a in ft/sec2):

·  V = liftoff velocity; a = constant A/C acceleration; t = takeoff time

·  V2 = 2aX, and V = at

·  X = V2 / 2a, and t = V / a.

·  X µ V2 / a.

·  t µ V / a.

These equations say:

·  Takeoff distance X is

o  directly proportional to the square of liftoff velocity V.

o  inversely proportional to constant aircraft acceleration a.

·  Takeoff time t is

o  directly proportional to liftoff velocity V.

o  inversely proportional to constant aircraft acceleration a.

Thus we may write:

·  X in feet and t in seconds.

·  Since ratios are used, we may express airspeed V in KTAS without bothering to convert to ft/sec

·  Acceleration drops significantly during the takeoff roll. However, we assume constant acceleration. Ratios minimize errors deriving from this assumption.

Effect of Changes in Gross Weight

When W1 ® W2:

·  Since V2 / V1 = Ö (W2 / W1), V22 / V12 = W2 / W1.

·  Since F = T (thrust) = m a = (W/g) a, a µ1/W, so.

·  Thus.

·  To summarize

V2 = V1 Ö (W2 / W1).

X2 = X1 (W2 / W1)2.

t2 = t1 (W2 / W1)1.5.

Effect of Changes in Field Elevation (Jet or Normally Aspirated Recip)

When s1 ® s2

·  Since V2 / V1 = Ö (s1 / s2), V22 / V12 = s1 / s2.

·  Since T = m a = (W/g) a, a µT, and assuming T µ air density s, .

·  Thus.

·  To summarize:

V2 = V1 Ö (s1 / s2).

X2 = X1 (s1 / s2)2.

t2 = t1 (s1 / s2)1.5.

Effect of Temperature Change at Fixed Pressure Altitude (Jet or Normally Aspirated Recip)

·  s = d / q, so for fixed d (fixed field elevation and altimeter setting), s µ 1 / q.

·  Thus, where TA is absolute temperature (C + 273 or F + 459.4).

·  Thus we may rewrite the equations

o  V2 = V1 Ö (s1 / s2)

o  X2 = X1 (s1 / s2)2

o  t2 = t1 (s1 / s2)1.5

·  as

V2 = V1 Ö (TA2 / TA1)

X2 = X1 (TA2 / TA1)2

o  t2 = t1 (TA2 / TA1)1.5

Effect of Wind on No-Wind Takeoff Distance, Time, and Liftoff Velocity

·  VW is the wind component parallel the runway:

o  Positive VW denotes a headwind.

o  Negative VW denotes a tailwind.

·  V2 = V1 – VW, where V2 is the liftoff ground speed (not airspeed)

·  a1 = a2 because wind does not change acceleration

·  Thus, .

·  To summarize:

o  V2 = V1 – VW.

o  X2 = X1 (1 - VW / V1)2.

o  t2 = t1 (1 - VW / V1).

Note that a tailwind or headwind DOES NOT AFFECT liftoff AIRSPEED

Example: if VLOF = 150 KEAS,

·  with a 25 KT headwind, liftoff groundspeed is 150 – 25 = 125 KEAS.

·  with a 25 KT tailwind, liftoff groundspeed is 150 + 25 = 175 KEAS.

Runway Gradient

For a runway that makes an angle Q with level ground, runway gradient RG is the slope of the runway, i.e. its rise over its run, expressed as a decimal fraction.

·  RG = DY / DX = tan Q.

·  A positive RG indicates an upslope.

·  A negative RG indicates a downslope.

·  Examples:

o  6’ rise per 1000’: RG = 6/1000 = 0.006

o  10 foot drop per 1000’: RG = -10/1000 = -0.01

RG and Effective Takeoff Thrust

In the diagram, the airplane is taking off “uphill”:

·  A force W sin Q opposes the thrust T exerted by the engine on takeoff.

·  Since tan Q » sin Q for small Q, W sin Q » W tan Q = W (RG)

·  Thus TEFFECIVE = T – W (RG) is the effective takeoff thrust.

o  If RG is positive, TEFFECTIVE is decreased by W (RG) (uphill takeoff).

o  IF RG is negative, TEFFECTIVE is increased W (RG) (downhill takeoff).

Thrust-to-Weight Ratio of an Airplane

For an airplane with available thrust T and weight W:

·  RT/W = T / W

·  Examples

o  T = 9,000# and W = 20,000#: RT/W = 9000/20000 = 0.45

o  T = 60,000# and W = 300,000#: RT/W = 60000/300000 = 0.20

o  W = 100,000# and RT/W = 0.15, T = (W) RT/W = (100000) 0.15 = 15,000#.

Note: Since RT/W = T / W

·  TLEVEL = (W) RT/W is effective thrust when talking off on a level runway.

·  W (RG) is the decrease (increase) in effective thrust when taking off on an upslope (downslope)

·  Thus TSLOPE = (W) RT/W – W (RG) is effective thrust when taking off on a sloped runway.

·  Effective takeoff thrust is decreased for an upslope runway and increased for a downslope runway.

Effect of Runway Gradient on Takeoff Distance X , Time t, and Velocity V

For an airplane with available thrust T1 taking off on a runway with gradient RG.

·  T1 = (W) RT/W is the effective takeoff thrust on a level runway

·  T2 = (W) RT/W – (W) RG is the effective takeoff thrust on a sloped runway

·  T1 / T2 = W RT/W / W (RT/W – RG) = RT/W / (RT/W – RG).

·  Since acceleration is proportional to effective thrust a1/a2 = RT/W / (RT/W – RG)

·  V2 = V1 since runway slope doesn’t affect liftoff speed.

·  Thus and

· 

·  To summarize:

o  V2 = V1.

o  X2 = X1 [RT/W / (RT/W – RG)].

o  t2 = t1 [RT/W / (RT/W – RG)].

Summary of Derived Equations

Liftoff (Ground) Speed / Takeoff Roll / Takeoff Time
Gross Weight:
W1 ® W2 / V2 = V1 Ö (W2 / W1) / X2 = X1 (W2 / W1)2 / t2 = t1 (W2 / W1)1.5
Density Ratio:
s1 ® s2 / V2 = V1 Ö (s1 / s2) / X2 = X1 (s1 / s2)2 / t2 = t1 (s1 / s2) 1.5
Temperature:
TA1 ® TA2 / V2 = V1 Ö (TA2 / TA1) / X2 = X1 (TA2 / TA1)2 / t2 = t1 (TA2 / TA1) 1.5
Headwind:
VW / V2 = V1 – VW / X2 = X1 (1 - VW / V1)2 / t2 = t1 (1 - VW / V1)
R/W Grad: RG &
T-W Ratio: RT/W / V2 = V1 / X2 = X1 [RT/W / (RT/W – RG)] / t2 = t1 [RT/W / (RT/W – RG)]

Rules of Thumb Deriving from these Equations

·  A 21% increase in gross weight results in a 10% increase in liftoff speed [1.1 = Ö1.21]

·  A 10% increase in gross weight gives

1.  a 5% increase in takeoff speed [Ö1.1 » 1.05]

2.  a 21% increase in takeoff distance [(1.1)2 = 1.21]

3.  a 15% increase in takeoff time [(1.1)1.5 » 1.15]

·  A headwind of 10% of liftoff speed (e.g. 15 KTS for 150 KT liftoff) gives

1.  a 19% reduction in takeoff distance [(1 - .1)2 = 0.81]

2.  a 10% reduction in takeoff time [ (1 – 0.1) = 0.9]

3.  a 10% reduction in liftoff ground speed [more relative wind]

·  A tailwind of 10% of liftoff speed (e.g. 15 KTS for 150 KT liftoff) gives

1.  a 21% increase in takeoff distance [(1 + .1)2 = 1.21]

2.  a 10% increase in takeoff time [ (1 + 0.1) = 1.10]

3.  a 10% increase in liftoff ground speed [less relative wind]

·  Shifting from a 10% headwind to a 10% tailwind (as above) increases

1.  takeoff distance by about 40% [19 + 21 = 40]

2.  takeoff time and liftoff ground speed by 20% [10 + 10 = 20]

Example Problems

Jet Transport SL SA: 25,000# thrust; 200,000# gross weight; calm wind; VLIFTOFF = 140 KTAS; 35 sec takeoff time; 4500’ takeoff roll; level runway

V1 = 140 KTAS/KEAS; t1 = 35 sec; X1 = 4500’; W1` = 200,000#; TA1 = 273+15 = 288o K; s1 = 1.0

Estimate takeoff airspeed, takeoff time, and takeoff roll at:

1. 300,000# gross: W2 = 300,000#,

V2 = V1 Ö(W2/W1) = 140 Ö(300000/200000) = 171.4642819 KTAS/KEAS

t2 = t1 (W2/W1)1.5 = 35 (300000/200000)1.5 = 64.29910575 sec

X2 = X1 (W2/W1)2 = 4500 (300000/200000)2 = 10,125.00000 ft

2. 6000 ft elevation: s2 = 0.83386

V2 = V1 Ö(s1/s2) = 140 Ö(1.0/0.83386) = 153.130303462 KTAS/KEAS

t2 = t1 (s1/s2)1.5 = 35 (1.0/0.83386)1.5 = 45.80023753 sec

X2 = X1 (s1/s2)2 = 4500 (1.0/0.83386)2 = 6440.883269 ft

3. 40o C: TA2 = 273 + 40 = 313o K

V2 = V1 Ö(TA2/TA1) = 140 Ö(313/288) = 145.9499534 KTAS/KEAS

t2 = t1 TA2/TA1)1.5 = 35 (313/288)1.5 = 39.654805047 sec

X2 = X1 (TA2/TA1)2 = 4500 (313/288)2 = 5315.15842 ft

4 (a) 25 KT headwind: Vw = 25

V2 = V1 – VW = 140 – 25 = 115 KTS ground speed (140 KTAS)

t2 = t1 (1 – Vw/V1) = 35 (1 – 25/140) = 28.75000000 sec

X2 = X1 (1 – Vw/V1)2 = 4500 (1 – 25/140)2 = 3036.352941 ft

4 (b) 20 KT Tailwind: Vw = - 20.

V2 = V1 – VW = 140 – (-20) = 160 KTS ground speed (140 KTAS)

t2 = t1 (1 – Vw/V1) = 35 (1 – (-20)/140) = 40.00000000 sec

X2 = X1 (1 – Vw/V1)2 = 4500 (1 – (-20)/140)2 = 5877.551021 ft

4 (c) Repeat (a) and (b) using the figure below:

(a) 25/140 = 0.189571429 » 0.19 (headwind ratio)
·  Left on x-axis to 0.19
·  Down to intersect curve
·  Right to y-axis: read 0.33
·  4500 - 0.33(4500) = 3015.0 ft
·  Compare 3036 (calculated) / b) 20/140 = 0.142857143 » 0.14 (tailwind ratio)
·  Right on x-axis to 0.14
·  Up to intersect curve
·  Left to y-axis: read 0.30
·  4500 + 0.30(4500) = 5850.0 ft
·  Compare 5878 (calculated)

5 (a) 2% runway upslope (= 2’ rise per 100’)

RG = 2/100 = 0.02; RT/W = T/W =25000/200000 = 0.125000000000

V2 = V1 = 140 KTAS/KEAS

t2 = t1 [RT/W / (RT/W – RG)] = 35 [0.125/(0.125 – 0.02)] = 41.66666665 sec

X2 = X1 [[RT/W / (RT/W – RG)] = 4500 [0.125/(0.125 – 0.02)] = 5357.142855 ft

5 (b) 1.5% downslope (= 1.5’ fall per 100’)

RG = -1.5/100 = -0.015

V2 = V1 = 140 KTAS/KEAS

t2 = t1 [RT/W / (RT/W – RG)] = 35 [0.125/(0.125 – [-0.015])] = 31.25000000 sec

X2 = X1 [RT/W / ( RT/W – RG)] = 4500 [0.125/(0.125 – [-0.015])] = 4017.857143 ft

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