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Landing braking
Description 
Airplanes use three mechanisms to brake during landing:
• Thrust reversal, which involves redirecting the engine thrust forward instead of backward.
• Wing spoilers, which increase the coefficient of drag by exposing a surface to the airflow.
• Conventional wheel brakes.
[1] Review of Thrust Reverser Mechanism used in Turbofan Jet Engine Aircraft, Mohd Anees Siddiqui, Md Shakibul Haq, International Journal of Engineering Research and Technology, Volume 6, Number 5 (2013), pp. 717-726, diagrams [2] Michael Fabry, F-GHXX Boeing 737-2A1(Adv) Some reverse thrust action during a very rainy day, (jetphotos.com) - center, left [3] N90024 American Airlines Airbus A319-115(WL), AIRCANADA087, (planespotters.net) - center, right
The image below shows two types of thrust reversers: the first one uses a deflector that is moved into the airflow exiting the engine, while the second one directly deflects the flow forward.
ID:(14476, 0)
Model
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Parameters
Variables
Calculations
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Calculations
Calculations
Variable 
Given Calculate 
Target : Equation To be used 
Equations
$ a_p = \displaystyle\frac{ F_p }{ m }$
a_p = F_p / m
$ F_W =\displaystyle\frac{1}{2} \rho S_p C_W v_L ^2$
F_W = rho * S_p * C_W * v ^2/2
$ s = \left( v_L - a_p \displaystyle\frac{ t }{2}\right) t $
s = ( v_L - a_p * t /2)* t
$ s_L = \left( v_L - a_p \displaystyle\frac{ t_L }{2}\right) t_L $
s_L = ( v_L - a_p * t_L /2)* t_L
$ \tau_p = \displaystyle\frac{ v_p }{ a_p }$
tau_p = v_p / a_p
$ t_L = \tau_p \arctan\left(\displaystyle\frac{ v_L }{ v_p }\right)$
t_L = tau_p * atan( v_L / v_p )
$ v = v_L - v_p \tan\left(\displaystyle\frac{ t }{ \tau_p }\right)$
v = v_L - v_p * tan( t / tau_p )
$ v_p = \sqrt{\displaystyle\frac{2 F_p }{ \rho S_p C_W } }$
v_p = sqrt( 2* F_p /( rho * S_p * C_W ))
ID:(15187, 0)
Resistance force
Equation
The resistance force ($F_W$) kann mit the density ($\rho$), the coefficient of resistance ($C_W$), the total object profile ($S_p$) und the speed with respect to the medium ($v$) entsprechend berechnet werden folgende Formel:
| $ F_W =\displaystyle\frac{1}{2} \rho S_p C_W v_L ^2$ |
| $ F_W =\displaystyle\frac{1}{2} \rho S_p C_W v ^2$ |
Similarly to how the equation for the lift force ($F_L$) was derived using the density ($\rho$), the coefficient of lift ($C_L$), the surface that generates lift ($S_w$), and the speed with respect to the medium ($v$)
| $ F_L =\displaystyle\frac{1}{2} \rho S_w C_L v ^2$ |
in this analogy, what corresponds to the surface that generates lift ($S_w$) will be equivalent to the total object profile ($S_p$) and the coefficient of lift ($C_L$) to the coefficient of resistance ($C_W$), thus the resistance force ($F_W$) is calculated:
| $ F_W =\displaystyle\frac{1}{2} \rho S_p C_W v ^2$ |
The drag coefficient is measured and, in turbulent flows over aerodynamic bodies, values are generally found around 0.4.
ID:(4418, 0)
Initial acceleration
Equation
At the beginning of takeoff, aerodynamic resistance, which depends on velocity, is minimal. Therefore, the maximum acceleration ($a_p$) is determined solely by the propulsion force ($F_p$) and the body mass ($m$):
| $ a_p = \displaystyle\frac{ F_p }{ m }$ |
As aerodynamic resistance starts to reduce the propulsion force, this initial acceleration will be the maximum possible.
ID:(14506, 0)
Maximum speed
Equation
The propulsion force ($F_p$) counteracts the resistance force ($F_W$) by generating velocity, which in turn increases the same resistance force, as described in the total object profile ($S_p$), the coefficient of resistance ($C_W$), the density ($\rho$), and the speed with respect to the medium ($v$) in
| $ F_W =\displaystyle\frac{1}{2} \rho S_p C_W v_L ^2$ |
This process continues to increase the velocity until the point where the propulsion force equals the resistance force, representing the maximum achievable speed.
By equating the propulsion force with the resistance force and solving for velocity, we obtain the maximum speed ($v_p$):
| $ v_p = \sqrt{\displaystyle\frac{2 F_p }{ \rho S_p C_W } }$ |
If we equate the propulsion force ($F_p$) with the resistance force ($F_W$) with the total object profile ($S_p$), the coefficient of resistance ($C_W$), the density ($\rho$), and the speed with respect to the medium ($v$) in
| $ F_W =\displaystyle\frac{1}{2} \rho S_p C_W v_L ^2$ |
we obtain, for a the maximum speed ($v_p$),
$F_p = \displaystyle\frac{1}{2} \rho S_w C_L v_p ^2$
which, when solved for the maximum velocity, results in
| $ v_p = \sqrt{\displaystyle\frac{2 F_p }{ \rho S_p C_W } }$ |
As aerodynamic resistance starts to reduce the propulsion force, this initial acceleration will be the maximum possible.
ID:(14507, 0)
Characteristic time
Equation
With the acceleration generated by the engines, represented as
| $ a_p = \displaystyle\frac{ F_p }{ m }$ |
and the maximum velocity associated with the resistance, described by
| $ v_p = \sqrt{\displaystyle\frac{2 F_p }{ \rho S_p C_W } }$ |
we can define a characteristic time using the following expression:
| $ \tau_p = \displaystyle\frac{ v_p }{ a_p }$ |
This time provides an estimate of the order of magnitude of the takeoff and landing process, which typically occurs within a few minutes.
ID:(14510, 0)
Landing speed
Equation
The equation to calculate the speed with respect to the medium ($v$) in the takeoff time ($t$) with the maximum acceleration ($a_p$) and the maximum speed ($v_p$) is as follows:
| $\displaystyle\frac{dv}{dt}=-a_p\left[1- \left(\displaystyle\frac{v}{v_p}\right)^2\right]$ |
Upon integration, it results in the takeoff/landing propulsion time ($\tau_p$) and the landing speed ($v_L$).
| $ v = v_L - v_p \tan\left(\displaystyle\frac{ t }{ \tau_p }\right)$ |
With the equation for the speed with respect to the medium ($v$) in the takeoff time ($t$) with the maximum acceleration ($a_p$) and the maximum speed ($v_p$):
| $\displaystyle\frac{dv}{dt}=-a_p\left[1- \left(\displaystyle\frac{v}{v_p}\right)^2\right]$ |
it can be integrated from an initial value of the landing speed ($v_L$)
$\displaystyle\int_{v_L}^v \displaystyle\frac{dv}{1 - v^2/v_p^2} = -\displaystyle\int_0^t dt a_p$
and with the definition of the takeoff/landing propulsion time ($\tau_p$)
| $ \tau_p = \displaystyle\frac{ v_p }{ a_p }$ |
the result is
| $ v = v_L - v_p \tan\left(\displaystyle\frac{ t }{ \tau_p }\right)$ |
At the beginning, when the time is much smaller than the characteristic time, the tangent can be replaced by its argument. This implies that the velocity primarily decreases due to the influence of the engines.
ID:(14511, 0)
Landing time
Equation
The equation for speed with respect to the medium ($v$) for an aircraft during landing is given with the landing speed ($v_L$), the maximum speed ($v_p$), the takeoff/landing propulsion time ($\tau_p$), and the takeoff time ($t$) as follows:
| $ v = v_L - v_p \tan\left(\displaystyle\frac{ t }{ \tau_p }\right)$ |
Therefore, the landing time ($t_L$) is calculated using this equation for the case when the velocity at that moment is zero. This translates to:
| $ t_L = \tau_p \arctan\left(\displaystyle\frac{ v_L }{ v_p }\right)$ |
With the equation speed with respect to the medium ($v$) using the landing speed ($v_L$), the maximum speed ($v_p$), the takeoff/landing propulsion time ($\tau_p$), and the takeoff time ($t$) as follows:
| $ v = v_L - v_p \tan\left(\displaystyle\frac{ t }{ \tau_p }\right)$ |
where in the takeoff time ($t$) is equal to the landing time ($t_L$), we have:
$v = v_L - v_p \tan\left(\displaystyle\frac{t_L}{\tau_p}\right)=0$
If we solve this equation for time, we obtain:
| $ t_L = \tau_p \arctan\left(\displaystyle\frac{ v_L }{ v_p }\right)$ |
ID:(14513, 0)
Path taken upon landing
Equation
Given that the speed with respect to the medium ($v$) during landing varies with respect to the takeoff time ($t$) with the landing speed ($v_L$), the maximum speed ($v_p$), and the takeoff/landing propulsion time ($\tau_p$) according to the equation:
| $ v = v_L - v_p \tan\left(\displaystyle\frac{ t }{ \tau_p }\right)$ |
we can calculate the distance traveled along the runway by integrating this equation over time:
| $ s = \left( v_L - a_p \displaystyle\frac{ t }{2}\right) t $ |
Since the speed with respect to the medium ($v$) during landing varies with respect to the takeoff time ($t$) with the landing speed ($v_L$), the maximum speed ($v_p$), and the takeoff/landing propulsion time ($\tau_p$) according to the equation:
| $ v = v_L - v_p \tan\left(\displaystyle\frac{ t }{ \tau_p }\right)$ |
it is equal to the path taken on the strip ($s$) as a function of the takeoff time ($t$).
We can integrate the equation:
$\displaystyle\frac{ds}{dt}=v_0-v_p\tan\left(\displaystyle\frac{t}{\tau_p}\right)$
Obtaining the path as:
$s = v_L t + \log(|\cos( t / \tau_p)|) v_p \tau_p$
If the takeoff time ($t$), the logarithmic factor can be expanded up to the third order, resulting in the landing path being equal to:
| $ s = \left( v_L - a_p \displaystyle\frac{ t }{2}\right) t $ |
The resulting equation is a third-order approximation of $t/\tau_p$, which means that aerodynamic aids for braking are significantly reduced compared to engine thrust reversal.
Furthermore, we can use the landing time to estimate the required runway length for landing.
ID:(14514, 0)
Necessary runway length
Equation
The landing concludes when the speed with respect to the medium ($v$) becomes zero, which implies that the time ($t$) equals the landing time ($t_L$), resulting in a reduction of the path taken on the strip ($s$) to the landing Distance ($s_L$). Therefore, with the maximum acceleration ($a_p$) and the landing speed ($v_L$), the following is obtained:
| $ s_L = \left( v_L - a_p \displaystyle\frac{ t_L }{2}\right) t_L $ |
ID:(15997, 0)