Verifying the risk of aircraft exceeding the declared takeoff distance known as F.A.R. takeoff runway length requirements through real simulation of Boeing 737-800 aircraft in standard conditions known as the ICAO Standard Atmosphere (ISA).

Initial data:

The analysis of measurements indicates that the normal distribution fits both takeoff phases, the ground run phase length and the air phase length. It is worth noting that together, normal distribution as well as the Gram-Charlier distribution laws were examined and both fit takeoff length of such aircraft category. However, due to the limitation, we are presenting the normal distribution case in this paper ( Figure 2 ). Fifty four (54) observations (data points) are used to establish the distribution laws of both takeoff phases, ground run and air section.

A takeoff with all engines running and precisely at the limit of the first phase (ground phase), the pilot performs the following actions [3] [4] :

Acknowledging that the time required for the aircraft to rotate is $t_{В1}=6$ seconds and understanding that the length for which the risk of 50% the aircraft overrun beyond the expected accelerate distance is the minimum length of the takeoff ground run, the critical length value of the takeoff first phase (the ground run) is determined using the expression (1), m:

$L_{TO-1Ф}^{cr}=\frac{{V^{\prime }}_{TO-1Ф}\times t_{В1}}{3.6}+\frac{V_r^2}{{3.6}^2\times 2\times a_{В1}}$, (1)

$L_{TO-1Ф}^{cr}=\frac{\left(265.23+277.86\right)\times 6}{2\times 3.6}+\frac{{265.23}^2}{{3.6}^2\times 2\times 2.08}=1759.4$

where $V_r$ (km/h) is the speed at which the rotation of the aircraft, during takeoff ground run, should be initiated. Upon FAA rules, this speed is set with respect to the stalling speed of the aircraft in the takeoff configuration $V_{st-вз}$ according to the following’s relationships:

$V_r=1.05V_{st-вз}=1.05\times 3.6\times \sqrt{\frac{2M_{TOF}g}{C_{y\max }\rho S_w}}$, (2)

$V_r=1.05\times 3.6\times \sqrt{\frac{2\times 79016\times 9.8066}{1.225\times 103.65\times 2.48}}=265.23\text{km}/\text{h};$

where:

$M_{TOF}$—the maximum take-off weight, kg;

$\rho$—air density;

$S_w$—aircraft wing area;

$C_{y\max }$—the maximum lift coefficient of the aircraft in takeoff configuration;

$t_{В1}$—the time required to rotate the aircraft.

Using formula (3) we determine the acceleration within the takeoff first phase, from the moment of movement initiation when the speed is zero and the runway longitudinal slope is zero (the angle ${\theta }^{\circ }=0$) until the aircraft lift-off when reaching $V_{lift}$.

$a_{В1}=\frac{1}{M_{TOF}}\left(T_{To}-\frac{1}{2}\rho V_r^2{S}_w\left(C_{DT}-C_{LTO}\left(f_{В1}+{\phi }_{В1}\right)\right)-M_{TOF}g\cos \theta \left(f_{В1}+{\phi }_{В1}\right)\right)$(3)

$\begin{array}{c}{a}_{В1}=\frac{1}{79016}(219469.29-\left(\frac{1}{2}\times 1.225\times {\left(\frac{265.23}{3.6}\right)}^2\times 103.65\left(0.512-2.48\left(0.0340+0.4803\right)\right)\right)\\ \begin{array}{c}\\ \end{array}-79016\times 9.81\times 1\times \left(0.0340+0.4803\right))\end{array}$

$a_{В1}=2.28\text{m}/{\text{s}}^{\text{2}}$

The aircraft lift coefficient in take-off configuration during ground roll $C_{LTO}$, is calculated using the known aerodynamic formula:

$C_{LTO}=\frac{2M_{TOF}}{\text{319}\text{.324}\times \rho \times S_w{C}_{LTO-\max }^2}=2.48$ (4)

$C_{LTO-\max }$—the predetermined maximum lift coefficient at take-off configuration and maximum angle of attack.

$C_{DT}$—the corresponding drag coefficient is:

$C_{DT}=C_{D0}+k_0{C}_{LTO}^2=0.512$ (5)

$T_{To}$—the thrust force (kg∙m/s²) required to achieve at least the takeoff safety speed $V_2$ that must be attained at 10.7 meters height above the end of the runway (above the threshold). Upon regulation $V_2=1.2V_{st-вз}$, so the minimum thrust force is determined by the expression: [5] - [7]

$T_{To}=\frac{1}{2}\rho V_2^2{S}_w{C}_{D0}+\left(\frac{{\left(M_{TOF}g\right)}^2}{0.5\rho V_2^2{S}_w}\right)\left(\frac{1}{\pi eAR}\right)$ (6)

$\begin{array}{c}{T}_{To}=0.5\times 1.225\times {\left(\frac{303.12}{3.6}\right)}^2\times 103.65\times 0.0268\\ +\left(\frac{{\left(79016\times 9.81\right)}^2}{0.5\times 1.225{\left(\frac{303.12}{3.6}\right)}^2\times 103.65}\right)\left(\frac{1}{3.1416\times 0.36\times 5.68}\right)\\ =219469.29\end{array}$

$AR$—is the wing aspect ratio, the wing span to its area.

$e=4.61\left(1-0.045{\left(\frac{u^2}{S_w}\right)}^{0.68}\right){\left(\cos {\Delta }_{LE}\right)}^{0.15}-3.1$, (7)

where $C_{D0}$—is the drag coefficient value when the lift force is zero;

$K_0$—the drag correction factor;

$e$—the Oswald wing efficiency factor;

$u$—the wing span;

$S_w$—the wing area;

${\Delta }_{LE}$—the wing chamfer angle (between the wing outer edge and the axis perpendicular to the vector of moving direction of the aircraft).

The values of the coefficient of friction ${\phi }_{В1}$ and the coefficient of rolling resistance $f_{В1}$ at the given speed $V$ are determined, respectively, to be 0.4803 and 0.034. The friction value $\phi$ used in the formula presented below is established based on FAA system of runway surface friction threshold to initiate runway surface rehabilitation. FAA sets this value to the order of 0.52. The coefficient of friction is determined according to the formula:

${\phi }_{В1}=\phi -{\beta }_{\phi }V$, (8)

the coefficient ${\beta }_{\phi }$ is tabulated in Appendix 2 [8] ;

$V$—the speed of the aircraft at the moment of takeoff km/h;

$f$—the calculated values of the rolling resistance coefficients are established according to dependence:

$f_{{}_{В1}}=f_{20}+K_{f}V$, (9)

$f_{20}$—the values of the rolling resistance coefficient at the aircraft speed tabulated in Appendix 3 [8] ;

$K_f$—This coefficient is taken from the note to Appendix 3 [8] .

The critical value of the length of the takeoff air section, takeoff second phase, with all engines operating from the point at which the aircraft lifts the ground to the point at which the aircraft crosses the threshold at an altitude 10.7 meters above the runway $L_{TO-2Ф}^{cr}$ (m) is calculated using the following relationship [9] :

$L_{TO-2Ф}^{cr}=\sqrt{\left(2\times R_{Toff-1}\times ТСH_{В}\right)-{\left(ТСH_{В}\right)}^2}$, (10)

$L_{TO-2Ф}^{cr}=\sqrt{\left(2\times 20199.3\times 10.7\right)-{\left(10.7\right)}^2}=657.38$

where $R_{Toff-1}$—the radius of the flight path curve from the aircraft liftoff and until it clears an obstacle equal to a height of 10.7 meters in accordance with the civil aviation flight safety standards, (see Figure 1 ) and determined by the formula [9] [10] , m:

$R_{Toff-1}=\frac{V_{stall}^2}{g\left(n_1-1\right)}=\frac{{252.6}^2}{g\left(1.322-1\right)}=20199.3$, (11)

where $n_1$—the coefficient of load factor, the ratio of the aircraft’s lift to its weight force. When the aircraft takeoff with all engines operating this coefficient is determined as follows:

$n_1=\frac{L}{W}=1.322$, (12)

Hence, the takeoff runway length in standard day ISA conditions $L.F.A.R{.}_{Toff}$(M) is calculated:

$L.F.A.R_{Toff}=1.15\left(L_{TO-1Ф}^{cr}+L_{TO-2Ф}^{cr}\right)=2779.3$

It is worth noting that our calculation shows almost the same value as that declared on the Boeing Airplane Characteristics for Airport Planning. 2779.3 Vs. 2792.00 meters.

During takeoff with all engines operating, the 15% (required by regulations) added to the critical length value reflects the sum of the permissible deviation of the takeoff operational length in both phases, ground run stretch (first phase) and air stretch to clear the obstacle of 10.7 meters (second phase) ( Figure 1 ). Since all operational speeds have a linear dependence with respect to the stall speed in the takeoff configuration, it is fair to calculate the permissible deviation of the operational length of the first and second phases, respectively, as follows:

$L_{TO-1Ф}^{Perm}=0.15L_{TO-1Ф}^{cr}=263.913$, (13)

$L_{TO-2Ф}^{Perm}=0.15L_{TO-2Ф}^{cr}=98.607$, (14)

Thus the operational estimated distance, for which the risk of the front wheels of the aircraft runoff over the calculated distance, from the starting point to the point where the aircraft lifts off the ground (when the aircraft becomes airborne), with all engines operating, should be determined as per the regulation rule, m:

$L_{TO-1Ф}=1.15L_{TO-1Ф}^{cr}=2023.3$

The standard deviation of the aircraft’s take-off critical length with all engines operating in the first phase (ground acceleration run), can be established by differentiating the formula of the critical length (1) by the speed of movement, the friction coefficient, the pilot’s reaction time and the regulated decision speed at its limit:

${\sigma }_{L_{TO-1Ф}^{cr}}=\sqrt{{\left(\frac{t_{ВР}}{\text{3}\text{.6}}\right)}^{\text{2}}{\sigma }_{{V^{\prime }}_{TO-1Ф}}^2+{\left(\frac{V_r}{a_{В1}}\right)}^{\text{2}}{\sigma }_{V_r}^{\text{2}}+{\left[\frac{V_r^2}{\text{2}{\left(\pm i+a_{В1}\right)}^{\text{2}}}\right]}^{\text{2}}{\sigma }_{{\phi }_{В1}}^{\text{2}}+{\left(\frac{{V^{\prime }}_{СР-ВВД}}{\text{3}\text{.6}}\right)}^{\text{2}}{\sigma }_{t_{ВР}}^{\text{2}}}$, (15)

$\begin{array}{c}{\sigma }_{L_{TO-1Ф}^{cr}}=\sqrt{{\left(\frac{6}{\text{3}\text{.6}}\right)}^{\text{2}}\times {4.27}^2+{\left(\frac{73.67}{2\times 2.08}\right)}^{\text{2}}\times {4.18}^2+{\left[\frac{\text{7}{3.67}^{\text{2}}}{\text{2}\times \text{2}{\text{.08}}^{\text{2}}}\right]}^{\text{2}}\times \text{0}{.02}^{\text{2}}+{\left(\frac{271.54}{\text{3}\text{.6}}\right)}^{\text{2}}\times {0.9}^2}\\ =100.839\end{array}$

where— ${\sigma }_{{V^{\prime }}_{TO-1Ф}}=4.27$, ${\sigma }_{V_r}=4.18$, are respectively, the standard deviations of the average speed of the aircraft during the rotation sub-phase, the speed of the aircraft at the moment the nose wheel gear is raised (m/h). These values are calculated using the relationship: ${\sigma }_V=\left(0.05\times V\right)+0.5$;

${\sigma }_{{\phi }_{В1}}=0.02$—the standard deviations of the coefficient of runway surface friction (pre-calculated);

${\sigma }_{t_{ВР}}=0.09$—the standard deviations of the pilot reaction time (pre-calculated);

As a result, the permissible coefficient of variation of the critical run length during takeoff in the first phase is determined according to the formula:

$C_v^{{}_{L_{ВВД-1Ф}^{КР}}}=\frac{{\sigma }_{L_{TO-1Ф}^{cr}}}{L_{TO-1Ф}^{cr}}=\frac{100.839}{1759.42}=0.057$, (16)

The coefficients of variation of the operational (permissible) and critical (minimum) lengths are comparable and equal $C_v^{{}_{L_{TO-1Ф}^{cr}}}=C_{\nu }^{{}_{L_{TO-1Ф}}}$ and the operational estimated distance as per regulation requirements ( $L_{TO-1Ф}$) is known. Hence, the standard deviation of the operational length of the first phase is determined from the expression, m:

${\sigma }_{L_{TO-1Ф}}=C_{\nu }^{{}_{L_{TO-1Ф}}}\times L_{TO-1Ф}^{Perm}=0.057\times 263.913=15.13$; (17)

The observation of the experimental data shows excellent agreement between the normal distribution and the empirical probability law, when analyzing the statistical data, in addition to small coefficients of asymmetry and eccentricity calculated based on Gram-Charlier distribution. So the normal distribution law is applicable for mathematical modeling of the risk assessment of an aircraft takeoff length of both phases. Hence, the risk of the aircraft going beyond the regulated operational takeoff run distance, under normal operating standard conditions (ISA), with all engines operating is calculated according to Equation (18) below:

$r_{TO-1Ф}=0.5-Ф\left(\frac{L_{TO-1Ф}-L_{TO-1Ф}^{cr}}{\sqrt{{\sigma }_{L_{TO-1Ф}}^2+{\left({\sigma }_{L_{TO-1Ф}^{cr}}\right)}^2}}\right)$, (18)

$\begin{array}{c}{r}_{TO-1Ф}=0.5-Ф\left(\frac{2023.3-1759.4}{\sqrt{{15.126}^2+{100.839}^2}}\right)=0.5-Ф\left(2.58823\right)\\ =0.5-0.49532778=0.00467\end{array}$

Thus, the risk of the aircraft overrun beyond the expected acceleration distance is 4 × 10⁻³ (4 incidents per 1000 operations).

By determining the risk of the aircraft exceeding the takeoff length of the first takeoff phase, we move to the second takeoff phase (the air phase) to calculate the risk of exceeding the expected horizontal length in order to clear the 10.7 meters obstacle when crossing the threshold.

The standard deviation of the critical length of the second phase during takeoff with all engines running can be determined by differentiating the formula of horizontal critical length of the second phase (5) by the average speed of the aircraft during the takeoff air phase:

${\sigma }_{L_{TO-2Ф}^{cr}}=\sqrt{5.128058\times V_{stall}\times {\sigma }_{V_{stall}}}$, (19)

${\sigma }_{L_{TO-2Ф}^{cr}}=\sqrt{5.128058\times 4.01\times \left(252.6/3.6\right)=}37.98$

Therefore, the permissible coefficient of variation of the critical length of the second phase is determined by the formula:

$C_v^{{}_{L_{ВВД-2Ф}^{КР}}}=\frac{{\sigma }_{L_{TO-2Ф}^{cr}}}{L_{TO-2Ф}^{cr}}=\frac{37.98}{657.38}=0.06$, (20)

At the boundary limit the variation coefficients of both, the operational (permissible) and critical parameters of the takeoff air phase length are equal to each other:

$C_v^{{}_{L_{ВВД-2Ф}^{КР}}}=C_{\nu }^{{}_{L_{TO-2Ф}}}$

and the permissible deviation of the operational length of the second phase is

$L_{TO-2Ф}^{Perm}=0.15L_{TO-2Ф}^{cr}=98.607$, (21)

So the standard deviation of the calculated distance along the trajectory of the air section becomes:

${\sigma }_{L_{TO-2Ф}}=C_{\nu }^{{}_{L_{TO-2Ф}}}\times L_{TO-2Ф}^{Perm}=0.06\times 98.606=5.7$

Thus, the quintile value (u) of Laplace function $Ф\left(u\right)$ of the takeoff horizontal length of the second phase becomes known as:

$U=\frac{L_{TO-2Ф}-L_{TO-2Ф}^{cr}}{\sqrt{{\sigma }_{L_{TO-2Ф}}^2+{\left({\sigma }_{L_{TO-2Ф}^{cr}}\right)}^2}}$

Then the risk of the aircraft exceeding the horizontal length of the takeoff air segment under normal operating standard conditions (ISA) with all engines operating is as follows:

$r_{TO-2Ф}=0.5-Ф\left(\frac{L_{TO-2Ф}-L_{TO-2Ф}^{cr}}{\sqrt{{\sigma }_{L_{TO-2Ф}}^2+{\left({\sigma }_{L_{TO-2Ф}^{cr}}\right)}^2}}\right)$, (22)

$\begin{array}{c}{r}_{TO-2Ф}=0.5-Ф\left(\frac{755.99-657.38}{\sqrt{{5.7}^2+{37.98}^2}}\right)=0.5-Ф\left(2.56777\right)\\ =0.5-0.49488168=0.0051\end{array}$

Thus, the risk of the aircraft exceeding the horizontal length of the takeoff air segment is 5 × 10⁻³ (5 incidents per 1000 operations).

Note that Laplace function $Ф\left(u\right)$ could be determined from Appendix 3 in Ref. [8] .

Finally, understanding that during takeoff, two consecutive risk situations are arising, where each situation has its own independent parameters, the value of total risk is determined via the formula:

$r_{TO}=r_{TO-1Ф}+r_{TO-2Ф}-r_{TO-1Ф}\times r_{TO-2Ф}=0.01$, (23)

Thus, the risk of the aircraft exceeding the declared takeoff distance (F.A.R. take off runway length) is 1 × 10⁻² (1 incident per 100 operations). It becomes obvious that a runway length of 2792 meters does not meet the required permissible risk value of 1 incident per one million operations 1 × 10⁻⁶.

By iteration, adding 350 meters to the takeoff distance (F.A.R. take off runway length), resulting in minimizing the risk of aircraft running off the runway or failing to clear the obstacle of 10.7 meters requirements during the takeoff operation by means of mathematical modeling. We’ll demonstrate it.

Using Equation (18), the recalculation of the risk of the Boeing 737-800 going beyond the regulated operational takeoff run distance, under normal operating standard conditions (ISA), with all engines operating is:

$\begin{array}{c}{r}_{TO-1Ф}=0.5-Ф\left(\frac{2278.33-1759.4}{\sqrt{{15.126}^2+{100.839}^2}}\right)=0.5-Ф\left(5.08904977\right)\\ =0.5-0.49999924=\text{0}\text{.00000076}\end{array}$

$r_{TO-1Ф}=7.6\times {10}^{-7}$

Using Equation (22), the recalculation of the risk of the Boeing 737-800 exceeding the horizontal length of the takeoff air segment under normal operating standard conditions (ISA) with all engines operating is:

$\begin{array}{c}{r}_{TO-2Ф}=0.5-Ф\left(\frac{851-657.38}{\sqrt{{5.7}^2+{37.98}^2}}\right)=0.5-Ф\left(5.041605\right)\\ =0.5-0.49999918=\text{0}\text{.00000082}\end{array}$

$r_{TO-2Ф}=8.2\times {10}^{-7}$

Therefore, the value of total risk is

$r_{TO}=r_{TO-1Ф}+r_{TO-2Ф}-r_{TO-1Ф}\times r_{TO-2Ф}=0.0000058$,

$r_{TO-2Ф}=5.8\times {10}^{-6}$