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
o 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
o V2 = V1 Ö (W2 / W1).
o X2 = X1 (W2 / W1)2.
o 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:
o V2 = V1 Ö (s1 / s2).
o X2 = X1 (s1 / s2)2.
o 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
o V2 = V1 Ö (TA2 / TA1)
o 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 TimeGross 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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