Ground Roll

The forces acting on an airplane during the ground roll portion of the takeoff are shown in Figure 7.2. From Newton’s second law of motion,

T – D – p(W – L) = MV

L Figure 7.2 Forces on an airplane during the ground roll.

is the coefficient of rolling friction, and a dot above the V denotes differentiation with respect to time, ц values can range from approximately 0.02 to 0.1, depending on the surface. The lower value corresponds to a hard, dry surface; the higher value might correspond to moderately tall grass.

Before rotation, the attitude of the airplane on the ground is constant and hence CL and CD are constant. After rotation, CL and CD increase, but still remain constant until lift off. The most direct means of solving Equation 7.1 is numerically, using a digital computer. This will require a table lookup or a curve fit for T as a function of D. After V is obtained as a function of time, it can then be numerically integrated to obtain s so that V will be known as a function of s.

A word of caution regarding CD is in order. Ground effect may reduce the induced drag significantly. Hence CD as a function of CL is less during the ground roll than it is in the air. In view of this, a fairly good approximation is to neglect the induced drag for calculating the total airplane drag during the ground roll, particularly for tricycle landing gears where the wing is nearly level during the ground roll.

A simple model that can be used to predict the relative effect of the ground on CD, can be obtained by replacing the wing by a simple horseshoe vortex. The span of the vortex is taken to equal 7t/4 times the span of the wing, b, since the vortex sheet shed from an elliptic wing can be shown to roll up into a pair of vortices spaced this distance apart. This horseshoe vortex system, together with the image system required to cancel vertical velocities at the ground, is shown in Figure 7.3. The ratio of the downwash induced at the midspan of the bound vortex, including the image system to the down – wash without the image, equals approximately the ratio of the induced drag in ground effect (IGE) to the induced drag, out-of-ground effect (OGE). Using the Biot-Savart law, this ratio becomes

Cn(IGE)’ = (16 hlb)2

Cd,(OGE) 1+(16hlb)1 4 ]

Figure 7.3 Simplified vortex model for calculating the effect of the ground on induced drag.

Equation 7.2 is plotted in Figure 7.4 where, for a typical value of h/b of 0.1, the induced drag IGE is seen to equal only 20% of the induced drag OGE for the same CL-

During the ground roll an increment to the parasite drag is required for airplanes with retractable landing gear. This increment can be estimated using the material presented in Chapter Four or on the basis of Figure 7.5. (Ref. 7.4). This figure presents the equivalent flatyjlate area, /, as a function of gross weight for three different types of landing gear.

Returning to Equation 7.1, let us apply a simple Euler method to the numerical integration of this equation. For illustrative purposes, a Boeing 747-100 will be used. This airplane is powered by four JT9D-7A turbofans having the rated takeoff thrust shown in Figure 6.36. This thrust at sea level for one engine is approximated closely by

Г = 46,100-46.7 V + 0.0467 V2

where T is in pounds and V is in feet per second.

Standard sea level conditions are chosen together with a gross weight of

733,0 lb (3260 kN). hlb for this airplane is approximately 0.08, giving an 86% reduction in the induced drag due to ground effect. Since takeoff is made with partially deflected flaps, a Cl of 1.0 will be assumed. Clean, the equivalent flat-plate area for this airplane is estimated to be 100 ft2 (9.3 m2). With gear

0 0.1 0.2 0.3

h

j

Figure 7.4 Reduction in CD, resulting from ground effect.

and flaps down, / is assumed to double, having a value of 200 ft2 (18.6 m2). (See Equation 3.46 and Figure 7.5.) These numbers are strictly estimates and may not agree with those used by the ntanufacturer. These numbers, together with other constants to be used for this example, are listed below.

W = 733,000 lb S = 5500 ft2 A = 6.69 e = 0.7 b = 196 ft H = 0.02

CD.(IGE)/CD,(OGE) = 0.14 / = 100 ft2 clean = 200 ft2 gear and flaps down

The numerical integration uses the following approximations.

" V(t + At) = V(t)+ V(t)At

s(/ + AO = s(0 + [V(f)+V(/ + A/)]y

Initially the airplane is at rest, so that V is calculated from

= 7.46 ft/s2

О 50 100 150 200 250 300 350 400 Takeoff weight/1000 lb Figure 7.5 Landing gear flat-plate area. (L. M. Nicolai, Fundamentals of Aircraft Design, L. M. Nicolai, 1975. Reprinted by permission of L. M. Nicolai.)

Choosing a time increment, At, of 0.1 sec for the numerical integration, at the next time step,

/=0.1 sec V = 0 + (7.46)(0.1)

= 0.746 fps

s =0 +(0 + 0.746X0. l)/2 = 0.0378 ft

With V now finite, the drag must be calculated. For this time step, D is negligible, being equal to only 0.171b. However, if this numerical integration is continued to, say, 32 sec, a distance of 3509 ft and a velocity of 210.3 fps are calculated. At this instant of time and for this velocity, the total thrust equals 153,377 lb and the lift and drag are calculated to be 289,200 lb and 13,1461b respectively. At this speed the drag is indeed significant. The in­stantaneous acceleration is found from Equation 7.1.

= 5.77 ft/sec2 (7.3)

Figure 7.6 presents the numerical solution of Equation 7.1 up to a velocity of 274.5 fps for the 747-100. This particular velocity is 10% above the stalling speed, assuming a of 1.8. In the next section, this particular value of V is chosen as the lift-off speed for calculating the flare distance over an obstacle.

Since CL and / are uncertain, Figure 7.7 was prepared to shovv the sensitivity of the ground roll distance to these parameters for a constant lift-off speed for the 747-100. s is seen to increase almost linearly with / while being rather insensitive to CL.

For preliminary design studies, an approximate method is frequently used tQ calculate the ground roll distance. The method is based on assuming that

Figure 7.6 Calculated velocity as a function of distance along the runway for a Boeing 747-100.

I___ I___ I___ I I I *’ I I’ I

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

Cl

Figure 7.7 Sensitivity of predicted ground roll distance to flat-plate area and CL for Boeing 747-100.

•i

the inverse of the acceleration is a linear function of the square of the velocity. There appears to be no real rational basis for this assumption other than the fact that the results to which it leads are reasonable. To begin, let

dV=Wdi= ydV _ 1 dV2 dt ds dt ds 2 ds

The term in parentheses can be identified as the reciprocal of the acceleration evaluated at V divided by V2. Thus, Equation 7.5 becomes

(7.6)

^ where a is an average acceleration evaluated at V/V2. For example, consider ‘_phe foregoing 747 example for a CL of 1.0, an / of 2000 ft2, and a V of 274.5 fps.

-7== 194.1 fps

V2

At this speed,

T = 155,1801b D = 11,2001b L = 247,100 lb

so that, from Equation 7.3,

d = 5.90 ft/sec2

The ground roll distance is then calculated from Equation 7.6 to be

= 247.52 S 2(5.9)

= 6386 ft

This result, by comparison to Figure 7.6, is seen to be within 1% of the more exact value obtained by the numerical methods.