TAKEOFF FUNDAMENTALS

Takeoff is a critical flight phase, one which constitutes only about one percent of total flight time, but which results in about to 1/5 (20%). of all accidents. In high performance, high speed airplanes, such accidents routinely involve serious damage and often result in fatalities. (Landing is an equally brief, even more critical flight phase, where about 1/4 (25%) of all accidents occur.)

We will examine five factors which affect three important takeoff parameters. The three takeoff parameters are:

  1. Takeoff distance X (distance covered from brake release to liftoff).
  2. Take off time t (elapsed time from brake release to liftoff).
  3. Takeoff speed V (liftoff speed).

The five factors are:

  1. Changes in Gross Weight.
  2. Changes in Density Altitude.
  3. Changes in Temperature.
  4. Runway Wind Changes.
  5. Variation of Runway Slope from Horizontal.

We also give a real world example of how to compute takeoff distance using a takeoff performance chart for a single engine high performance Navy jet airplane. Finally, we briefly discuss the importance of a takeoff acceleration check and the hazard associated with premature rotation.

1. Theoretical Effects of Changes in Gross Weight, Density Altitude, Wind,

and Runway Slope on Liftoff Speed, Takeoff Distance, and Takeoff Time

During takeoff, thrust overcomes drag and tire friction forces to accelerate the airplane. We first examine equations which allow predicting takeoff roll, speed, and time changes when gross weight, density altitude, temperature, runway slope, or the headwind component changes

An increase in gross weight, temperature, or runway upslope, or a decrease in density altitude or headwind component, causes an increase in one or more of takeoff speed, distance, and time. Sometimes the increase can be drastic, especially when rising temperatures and higher takeoff elevations (density altitudes) lead to significantly lower air density.

For example, later we will discuss a single engine jet airplane whose takeoff roll (at fixed gross weight) increases from 2500’ to 10500’ when runway temperature rises from 0o to 40o C, pressure altitude decreases from sea level to 2000’, and headwind decreases from 25 kts to 0 kts. This is an increase of 420% in takeoff distance.

Implications of Equations for Rectilinear Motion. From basic physics, we know that

VF2 = VI2 + 2aX, and

VF = VI + at,

where VI and VF respectively are initial and final velocity in ft/sec; a is acceleration in ft/sec2, assumed to be constant (a naïve but useful assumption); X is distance traversed in feet; and t is time in seconds.

Applied to airplane takeoff, VF is liftoff velocity, VI = 0, X is takeoff roll, and we have

VF2 = 2aX, and VF = at, or

X = VF2 / 2a, and t = VF / a.

These equations tell us that:

  • Takeoff roll X is directly proportional to the square of liftoff speed and inversely proportional to acceleration:

X  V2 / a.

  • Takeoff time t is directly proportional to liftoff speed and inversely proportional to acceleration.

t  V / a.

Suppose a change in some factor such as gross weight or density altitude causes a change in liftoff speed and aircraft takeoff acceleration. Let X1, t1, and V1 be takeoff parameters associated with the initial situation, and X2, t2 and V2 be parameters associated with the new situation. Then from the foregoing we may conclude:

Note: since these identities involve ratios, we may use airspeed in knots without bothering to convert to ft/sec. Also, since we assume constant acceleration in both cases (when actually acceleration decreases significantly during the takeoff roll), the error introduced is not as great as one might expect.

Effect of Changes in Gross Weight. Suppose gross weight changes from W1 to W2. Then, as we already know, respective liftoff speeds V1 and V2 are given by

V2 / V1 =  (W2 / W1), or V22 / V12 = W2 / W1.

Also, from F = m a = (W/g) a, we see that at takeoff thrust F, acceleration is inversely proportional to gross weight. We may conclude that the respective accelerations a1 and a2 corresponding W1 and W2 are given by . (The equation says the acceleration is inversely proportional to weight.) Thus

.

That is:

  • V2 = V1 (W2 / W1).
  • X2 = X1 (W2 / W1)2.
  • t2 = t1 (W2 / W1)3/2.

Expressed in English, these equations say that if takeoff gross weight changes from W1 to W2:

  • Liftoff speed changes as the square root of (W2 / W1).
  • Takeoff roll changes as the square of (W2 / W1).
  • Takeoff time changes as (W2 / W1) raised to the 3/2 = 1.5 power.

Effect of Changes in Field Elevation (Jet or Normally Aspirated Prop). Suppose density ratio corresponding to field elevation changes from 1 to 2. Then, as we already know, respective liftoff speeds V1 and V2 are given by

V2 / V1 =  (1 / 2), or V22 / V12 = 1 / 2.

Also, from F = m a = (W/g) a, we see that at takeoff gross weight W, acceleration is directly proportional to thrust F. Assuming thrust is proportional to density ratio for a jet or normally aspirated prop aircraft, we may write . (The equation says that acceleration is inversely proportional to density ratio.) Thus

.

That is, for a turbojet or normally aspirated reciprocating prop airplane:

  • V2 = V1 (1 / 2).
  • X2 = X1 (1 / 2)2.
  • t2 = t1 (1 / 2)1.5.

Expressed in English, these equations say that if density ratio changes from 1 to 2:

  • Liftoff speed changes as the square root of (1 / 2).
  • Takeoff roll changes as the square of (1 / 2).
  • Takeoff time changes as (1 / 2) raised to the 1.5 power.

Effect of Temperature Change at Fixed Pressure Altitude(Jet or Normally Aspirated Prop). Air density is a function of pressure and temperature, as implied by the well-known relation  =  / . That is, for a fixed pressure altitude (fixed field elevation and altimeter setting),  1 / . Thus the equations given previously for density altitude change may be rewritten to reflect change in temperature at fixed static pressure as follows:

  • V2 = V1 (2 / 1).
  • X2 = X1 (2 / 1)2.
  • t2 = t1 (2 / 1)1.5.

Recall that temperature ratio  must be calculated using absolute temperatures. Let TA represent absolute temperature, with TA0 denoting SL temperature in a standard atmosphere. (To calculate TA, add 273 to temperature in degrees Celsius, or 460 to temperature in degrees Fahrenheit.) Then 2 = TA2 / TA0, and 1 = TA1 / TA0, so 2 / 1 = TA2 / TA1. Thus the above equations may be rewritten:

  • V2 = V1 (TA2 / TA1).
  • X2 = X1 (TA2 / TA1)2.
  • t2 = t1 (TA2 / TA1)1.5.

Expressed in English, these equations say that if absolute temperature changes from TA1 to TA2:

  • Liftoff speed changes as the square root of (TA2 / TA1).
  • Takeoff roll changes as the square of (TA2 / TA1).
  • Takeoff time changes as (TA2 / TA1) raised to the 1.5 power.

Effect of Headwind. The effect of a headwind component on takeoff is to decrease the ground speed of the aircraft at liftoff. An excellent illustration of this phenomenon is the “ground” speed and airspeed of a jet airplane being launched from an aircraft carrier. The captain accelerates the ship and turns it into the existing sea wind to get about 35 KTS of wind across the deck. Then an airplane which flies off the bow at 150 KIAS is only moving 150 – 35 = 115 KTS relative to the deck of the ship. That is, the catapult only has to accelerate the plane to 115 KTS in the space of 200’, not 150 KTS. (Note: there have been a few launches of jet airplanes from aircraft carriers at anchor, and these rarities are celebrated events in Naval Aviation lore, with the pilots involved accorded significant bragging rights.)

Let VW be the wind component parallel the runway, with a positive value denoting a headwind, and a negative value denoting a tailwind. We may conclude that

V2 = V1 – VW

with acceleration unchanged (i.e., a1 = a2). Thus

.

That is:

  • V2 = V1 – VW.
  • X2 = X1 (1 - VW / V1)2.
  • t2 = t1 (1 - VW / V1).

Expressed in English, these equations say that if takeoff gross weight changes from W1 to W2:

  • Liftoff ground speed changes by the headwind component VW.
  • Takeoff roll changes as the square of (1 - VW / V1)2.
  • Takeoff time changes as (1 - VW / V1).

Note: the decrease/increase in takeoff roll due to headwind/tailwind also applies to landing rollouts.

Note: Let (x, y) be a point on the curve depicted in Figure 7.1. Then y / 100 = x2 if x  0.0, and y / 100 = -x2 if x < 0. That is, the curve is parabolic, but this fact is obscured by different x and y scales. Also, numbers on the x-axis are decimal fractions of the takeoff (or landing) speed, while numbers on the y-axis are percentage changes in takeoff or landing distance, i.e., decimal fractions multiplied by 100.

Effect of Runway Slope or Gradient. The gradient G of a slope is just its rise over its run, expressed as a decimal fraction.

Let the Cartesian x-axis lie tangent to the earth’s surface, so that the y-axis is perpendicular to the earth. Then for a runway that has a non-zero gradient (upslope or downslope), the gradient  is just y / x, as illustrated below.

As illustrated in the above figure, for an upslope, there is an additional weight component W sin  which thrust must overcome, effectively decreasing thrust (the accelerating force) on takeoff. (If y is negative, the runway has a downslope, and thrust is increased by W sin ).

Since the sin  tan  for small , we may write sin  tan  = y / x = RG, where RG is runway gradient, so the additional force to be overcome is just W sin  W tan  = W (y / x) =W (RG). That is, a runway gradient changes effective takeoff thrust of an aircraft by W (RG), where W is takeoff gross weight. RG must be expressed as a decimal fraction, as is conventional. A positive value for RG denotes an upslope, and a negative value indicates a downslope. If the runway gradient is positive (uphill), effective thrust is decreased by W (RG); otherwise (downhill or negative gradient) effective thrust is increased by W (RG).

Let RT/W be the thrust to weight ratio for takeoff thrust. For example, if an airplane weighs 20,000# and has 9,000# takeoff thrust, RT/W = 9,000 / 20000 = 0.45. RT/W for high performance jet fighters often approaches 0.75, while RT/W for an airliner might be well below 0.5. Of course, W RT/W gives takeoff thrust in pounds for an airplane at gross weight W. For example, if W = 300,000# and RT/W = 0.20, then takeoff thrust is just W RT/W = 300,000# (0.20) = 60,000#.

Since takeoff thrust is W RT/W, a runway gradient of RG causes effective takeoff thrust (TR)1 = W RT/W to become (TR)2 = W RT/W - W (RG) = W (RT/W – RG), with both RT/W and RG expressed as decimal fractions. That is,

(TR)1 / (TR)2 = W RT/W / W (RT/W – RG) = RT/W / (RT/W – RG).

Now recall the relationships that express takeoff distance and time ratios. In the case of a runway gradient, liftoff velocity does not change since gross weight does not change. However, acceleration is changed by a factor of (TR)1 / (TR)2, since takeoff thrust TR = F = (W/g) a, with W/g constant. That is, V2 = V1, W2 = W1, and a1 /a2 = (TR)1 / (TR)2 = RT/W / (RT/W – RG). Thus:

To sum up, for runway gradient RG expressed as a decimal fraction (with a negative RG meaning a downslope):

  • V2 = V1.
  • X2 = X1 [RT/W / (RT/W – RG)].
  • t2 = t1 [RT/W / (RT/W – RG)].

Expressed in English, these equations say that for runway gradient RG and thrust to weight ratio RT/W:

  • Liftoff speed remains unchanged.
  • Takeoff roll and takeoff time change proportional to the ratio RT/W / (RT/W – RG).

Important note: runway slope affects takeoff distance significantly more in an airplane with low thrust-weigh ratio (such as an airliner) than in an airplane with high thrust-weight ratio (such as a jet fighter taking off with combat thrust.) This is because the ratio RT/W / (RT/W – RG), for fixed RG, becomes larger as RT/W becomes smaller.

Summary of Theoretical Effects of Gross Weight, Elevation, Temperature, Wind, and Runway Slope Changes. The foregoing results are summarized in the following table. Remember that the results for elevation and temperature changes apply to jets and normally aspirated props, but not to supercharged props. Also, neither one of these two results takes into account the very significant change in jet engine thrust as temperature and/or air density change. As a result, these two equations do not predict takeoff parameter changes as accurately as the other three.

Liftoff (Ground) Speed / Takeoff Roll / Takeoff Time
Gross Weight:
W1 W2 / V2 = V1 (W2 / W1) / X2 = X1 (W2 / W1)2 / t2 = t1 (W2 / W1)3/2
Density Ratio:
12 / V2 = V1 (1 / 2) / X2 = X1 (1 / 2)2 / t2 = t1 (1 / 2)3/2
Temperature:
TA1 TA2 / V2 = V1 (TA2 / TA1) / X2 = X1 (TA2 / TA1)2 / t2 = t1 (TA2 / TA1)3/2
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)]

Takeoff performance “rules of thumb” may be adduced from the above equations.. It is easy to verify their validity from the equations. [I have put hints in brackets to help you see how to do this.]

  • A 21% increase in gross weight results in a 10% increase in liftoff speed [1.1 = 1.21]
  • A 10% increase in gross weight gives
  1. a 5% increase in takeoff speed [1.1  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 [obvious]
  • 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 [obvious]
  • 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]

Factors Ignored by Foregoing Ratio Estimations. Several important factors are not taken into consideration by the theoretical approach above, which assumes constant acceleration on takeoff:

  1. The net accelerating force (Thrust – Drag– Friction) is not constant.
  2. Thrust drops during the runway roll for a jet engine because the air velocity at the engine intake increases faster than at the engine exhaust point.
  3. Parasite drag increases as a function of V2.
  4. After rotation, induced drag increases due to increased AOA.
  5. As lift develops, normal force on tires decreases, decreasing friction.
  6. As a consequence of 1-5 above, takeoff acceleration is not constant, although we assumed it is constant.

Figures 7.3 and graphs typical takeoff thrust, drag, friction, and net acceleration forces vs. airspeed for a B767. Figure 7.4 uses the net acceleration force of Figure 7.3 to graph typical B767 takeoff acceleration vs. airspeed.

Figure 7.3. Changes to Thrust, Drag, and Friction Forces during B767 Takeoff Roll

Figure 7.4. B767 Takeoff Acceleration Profile

Inaccuracies occur when one assumes constant acceleration during takeoff roll computation. In our equations, these inaccuracies are smaller because ratios are used to estimate changes in takeoff time, distance, and speed as factors such as gross weight, density, temperature, and pressure change.

2.Example Problems

Weight Change: An airliner at 200,000# gross takes 32 seconds and 4000’ to get airborne at 140 KTAS. At 300,000# gross with other conditions unchanged, find takeoff time,speed, and distance.

t2 = t1 (W2 / W1)1.5 = 32 (300,000 / 200,000)3/2 = 32 (1.5)1.5 = 58.78775382 sec

V2 = V1 (W2 / W1) = 140  (300,000 / 200,000) = 140 1.5 = 171 KTAS

X2 = X1(W2 / W1)2 = 4000 (300,000 / 200,000)2 = 4000 (1.5)2 = 9000.0000000 feet

Elevation Change. An airliner at SL on a standard day takes 32 seconds and 4000’ to get airborne at 140 KTAS. At 7000’ elevation with other conditions unchanged, find takeoff time, speed, and distance.

7000 = 0.81064

t2 = t1 (σ1 / σ 2)1.5 = 32 (1.0 / 0.81064)1.5 = 43.84377434 sec

V2 = V1 (σ1 / σ 2) = 140 1.0 / 0.81064) = 155.4941378 KIAS

X2 = X1 (σ1 / σ 2)2 = 4000 (1.0 / 0.81064)2 = 6087.008832 feet

Temperature Change.A jet fighter at SL takes 30 seconds and 4100’ to get airborne at 150 KTAS on a day when the temperature is 0o C. If the temperature increases to 40o C with other conditions unchanged, find takeoff time, speed, and distance.

TA2 = (40 + 273) = 313; TA1 = (0 + 273)

t2 = t1 (TA2 / TA1)1.5 = 30 (313 / 273)1.5 = 36.82932768 sec

V2 = V1 (TA2 / TA1) = 150 313 / 273) = 160.6135215w KIAS

X2 = X1 TA2 / TA12)2 = 4100 (313 / 273)2 = 5389.844633 feet

Wind Change. An airliner under no wind conditions takes 32 seconds and 4000’ to get airborne at 140 KTAS. Then with 30 KT headwind and other conditions unchanged, find takeoff time, speed, and distance.

t2 = t1 (1 – VW/V1) = 32 (1 – 30/140) = 25.14285714 sec

V2 = V1 - VW) = 140 - 30 = 110 KTAS

X2 = X1 (1 – Vw / V1)2 = 4000 (1 – 30/140)2 = 2469.38755 feet

If there is 25 KTS of tailwind, then takeoff time, ground speed and distance are

t2 = t1 (1 – VW/V1) = 32 (1 – (-25)/140) = 37.71428573 sec

V2 = V1 - VW) = 140 - (-25) = 165 KTAS

X2 = X1 (1 – Vw / V1)2 = 4000 (1 – (-25)/140)2 = 5556.122452 feet

Note that the liftoff speed IAS is not affected by headwind or tailwind; only the ground speed is changed.

Sometimes a chart (or table) is used to determine the percent change in takeoff or landing distances for a given headwind or tailwind. An example of such a chart is given in Figure 7.1.

Figure 7.1. Effect of Headwind / Tailwind Components on Takeoff and Landing Distance.

Previous Wind Change Problem Solved Graphically. An airliner under no wind conditions takes 32 seconds and 4000’ to get airborne at 140 KTAS. Then with 30 KTS headwind and other conditions unchanged, find takeoff time, distance. Also find the takeoff distance for a 25 KT tailwind.

30/140 = 0.21 (rounded) is the fractional relation of wind speed to liftoff speed. From the origin, proceed left on the x-axis to 0.21, then down to intersect the curve. Read 38% decrease in takeoff distance on the y-axis. 4000 – 0.38 (4000) = 2480 feet. This compares to 2469 feet computed by the equation.

-25/140 = -18% (rounded).is the fractional relation of wind speed to liftoff speed. From the origin, proceed right on the x-axis to 0.18, then up to intersect the curve. Read 39% increase in the takeoff distance on the y-axis. 1.39 (4000) = 5560 feet. This compares to 5556 feet computed by the equation.

Runway Slope. A 300,000# airliner with 90,000# takeoff thrust takes 32 seconds and 4000’ to get airborne at 140 KTAS on a level runway. With other conditions unchanged, find takeoff time and distance with a runway upslope of 2% and a runway downslope of 5%.

Thrust-weight ratio RT/W = 90,000/300,000 = 0.30.

With a 2% upslope, (RG = 0.02:

t2 = t1 [RT/W / (RT/W – RG)] = 32 [0.3 / (0.3 – 0.02)] = 34.28571427 sec

X2 = X1 [RT/W / (RT/W – RG)] = 4000 [0.3 / (0.3 – 0.02)] = 4285.714284 feet

Liftoff speed remains 140 KTAS.

With a 5% downslope, (RG = -0.05):

t2 = t1 [RT/W / (RT/W – RG)] = 32 [0.3 / (0.3 + 0.05))] =27.42857143 sec

X2 = X1 [RT/W / (RT/W – RG)] = 4000 [0.3 / (0.3 + 0.05)] = 3428.571428 feet

Liftoff speed remains 140 KTAS.

3. Real World Takeoff Distance Example

In contrast to our theoretical models, real world empirical data is used to construct takeoff charts which pilots (or computers) typically use to compute takeoff distance for given gross weight, temperature, density altitude, and headwind/tailwind component.

Navy RF8 Crusader Takeoff Distance Chart. The Takeoff Distance Chart for a U.S. Navy RF8 Crusader (Photo Reconnaissance version) using military power (as opposed to combat power, or afterburner) is given on the next page. While the F8 is an older supersonic aircraft (capable of exceeding Mach 1.8 and still being flown off ship by the French Navy as of November 1999), the methodology used to compute takeoff distance is typical and not outdated. To use the chart:

  1. Determine runway ambient temperature and headwind (or tailwind) component, field pressure altitude, and gross weight.
  2. At the lower left part of the chart, locate temperature on the vertical scale, and proceed horizontally left to intersect the appropriate curved pressure altitude line, interpolating as required.
  3. From this first intersection point, proceed vertically up to intersect the appropriate straight gross weight lines in the upper left part of the chart, interpolating as required.
  4. From this second intersection point, proceed horizontally to the right to intersect the vertical zero wind component line in the upper right part of the chart.
  5. From this third intersection point, proceed parallel the curved headwind/tailwind lines until intersecting the vertical line corresponding to headwind (or tailwind) component.
  6. From this fourth intersection point, proceed horizontally to the right to read takeoff distance.

Practice Calculations: Determine takeoff roll for 0o C, 25 kts of headwind, SL pressure altitude, and 28,000# gross. Also determine takeoff roll at the same gross weight for 40o C, calm wind, 2000’ pressure altitude.

The following table contains information extracted from the F8 Takeoff Chart. It answers the two problems posed above, and also compares actual takeoff distances to theoretical takeoff distances computed using previously derived equations. The distance “errors” using the equations are not always small, suggesting that the model used to formulate the equations may be somewhat naïve for predicting F8 Crusader takeoff performance. This is most noticeable in density/temperature changes, where the models do not take account of resultant, often large changes in jet thrust.

Note: Very wide variations in takeoff roll depending on takeoff conditions are reflected in the table. In particular, changes in pressure altitude and ambient temperature cause large changes in takeoff roll. The table figures should alert you to the fact that accurate takeoff performance calculations are crucial to flight safety when operating jet aircraft.

Runway Conditions / Chart Roll (ft) / Theoretical Takeoff Roll (ft) / % Diff
28000#, 40o C, 0 kts HWind, 2000 PA / 10500
28000#, 0o C, 0 kts HWind, 0 PA / 4100
28000#, 40o C, 0 kts HWind, 0 PA / 8800 / 4100 (313 / 273)2 = 5389 / -38.8%
28000#, 0o C, 0 kts HWind, 2000 PA / 4900 / 4100 (1.0 / 0.9428)2 = 4612 / -5.3 %
28000#, 0o C, 25 kts HWind, 0 PA / 2850 / 4100 (1 – 25/150)2 = 2847 / 0%
28000#, 40o C, 25 kts HWind, 2000 PA / 8000 / 10500 (1 – 25/154.5)2 =7376 / -7.8%
24000#, 0o C, 0kts HWind, 2000 PA / 3600 / 4900 (24,000 / 28.000)2 = 3600 / 0%
24000#, 0o C, 0 kts HWind, 0 PA / 2800 / 4100 (24,000 / 28.000)2 = 3012 / +7.5 %
24000#, 0o C, 0 kts HWind, 4000 PA / 3900 / 2800 (1.0 / 0.8881)2 = 3550 / -9.1 %