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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)
Resistance force
Equation
Analogously to the lift force, a pressure difference is generated at the front and rear of an object, resulting in a drag force. This drag force depends on the surface area facing the flow $S_p$, the velocity $v$, and the drag coefficient $C_W$, given by:
 $ F_W =\displaystyle\frac{1}{2} \rho S_p C_W v ^2$ |
The drag coefficient is measured and typically yields values around 0.4 for turbulent flows over aerodynamic bodies.
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 mass of Object ($m$):
| $ 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 ^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 ^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)
Acceleration upon landing
Equation
In essence, aircraft use propulsion systems to achieve braking during landing. To this the propulsion force ($F_p$), we add the resistance force ($F_W$), as described in the equation along with the density ($\rho$), the total object profile ($S_p$), the coefficient of resistance ($C_W$), and the speed with respect to the medium ($v$)
| $ F_W =\displaystyle\frac{1}{2} \rho S_p C_W v ^2$ |
This way, the total force equals the product of the mass of Object ($m$) and the acceleration ($a$), which can be expressed as the variation of the speed with respect to the medium ($v$) in terms of time ($t$), as mentioned in the label
| $ m \displaystyle\frac{dv}{dt} = - F_p - \displaystyle\frac{1}{2} \rho S_p C_W v ^2$ |
The total force is the propulsion force ($F_p$) plus the resistance force ($F_W$), which is calculated with the density ($\rho$), the total object profile ($S_p$), the coefficient of resistance ($C_W$), and the speed with respect to the medium ($v$) through
| $ F_W =\displaystyle\frac{1}{2} \rho S_p C_W v ^2$ |
resulting in the expression
$F = - F_p - \displaystyle\frac{1}{2} \rho S_w C_L v ^2$
Since the total force equals the mass of Object ($m$) multiplied by the acceleration ($a$), and the latter represents the change in the speed with respect to the medium ($v$) with respect to time ($t$), we can write
$F = m a = m \displaystyle\frac{dv}{dt}$
This leads us to the differential equation that describes the system's behavior:
| $ m \displaystyle\frac{dv}{dt} = - F_p - \displaystyle\frac{1}{2} \rho S_p C_W v ^2$ |
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ID:(14512, 0)
Landing speed equation
Equation
The equation for an airplane taking off with speed with respect to the medium ($v$) can be rewritten as follows when it takes off with the density ($\rho$), the total object profile ($S_p$), the coefficient of resistance ($C_W$), the mass of Object ($m$), the time ($t$), and the propulsion force ($F_p$):
| $ m \displaystyle\frac{dv}{dt} = - F_p - \displaystyle\frac{1}{2} \rho S_p C_W v ^2$ |
It can be rewritten with the maximum acceleration ($a_p$) and the maximum speed ($v_p$) as:
| $\displaystyle\frac{dv}{dt}=-a_p\left[1- \left(\displaystyle\frac{v}{v_p}\right)^2\right]$ |
The equation for an airplane taking off with speed with respect to the medium ($v$) can be rewritten as follows when it takes off with the density ($\rho$), the total object profile ($S_p$), the coefficient of resistance ($C_W$), the mass of Object ($m$), the time ($t$), and the propulsion force ($F_p$):
| $ m \displaystyle\frac{dv}{dt} = - F_p - \displaystyle\frac{1}{2} \rho S_p C_W v ^2$ |
This can be expressed as:
| $ a_p = \displaystyle\frac{ F_p }{ m }$ |
and the maximum speed ($v_p$)
| $ v_p = \sqrt{\displaystyle\frac{2 F_p }{ \rho S_p C_W } }$ |
as follows:
| $\displaystyle\frac{dv}{dt}=-a_p\left[1- \left(\displaystyle\frac{v}{v_p}\right)^2\right]$ |
ID:(15159, 0)
Landing speed
Equation
The equation to calculate the speed with respect to the medium ($v$) in the 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 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 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 time ($t$) as follows:
| $ v = v_L - v_p \tan\left(\displaystyle\frac{ t }{ \tau_p }\right)$ |
where in the 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 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 - v_p \displaystyle\frac{ t }{2 \tau_p }\right) t $ |
Since the speed with respect to the medium ($v$) during landing varies with respect to the 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 ($l$) as a function of the 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 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 - v_p \displaystyle\frac{ t }{2 \tau_p }\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)
Landing and abort speed ($V1$)
Equation
As the path taken on the strip ($l$) depends on the landing speed ($v_L$), the maximum speed ($v_p$), the takeoff/landing propulsion time ($\tau_p$), and the time ($t$) according to the following equation:
| $ s = \left( v_L - v_p \displaystyle\frac{ t }{2 \tau_p }\right) t $ |
it can be evaluated for the landing time ($t_L$), which makes the path taken on the strip ($l$) correspond to the required runway length the maximum speed ($v_p$). In the case of takeoff and not landing, the maximum speed ($v_p$) corresponds to the critical speed $V1$ ($V1$).
If we solve the equation for the braking process, whether for landing or emergency braking, we obtain the landing speed ($v_L$).
| $ v_L = \sqrt{2 a_p l }$ |
As the path taken on the strip ($l$) depends on the landing speed ($v_L$), the maximum speed ($v_p$), the takeoff/landing propulsion time ($\tau_p$), and the time ($t$) according to the following equation:
| $ s = \left( v_L - v_p \displaystyle\frac{ t }{2 \tau_p }\right) t $ |
and the landing time ($t_L$) is given by:
| $ t_L = \tau_p \arctan\left(\displaystyle\frac{ v_L }{ v_p }\right)$ |
therefore:
$s=\left(v_L-v_p\displaystyle\frac{1}{2}\arctan(v_L/v_p)\right)\tau_p\arctan(v_L/v_p)$
In the limit $v_L\ll v_p$, we have:
$s=\displaystyle\frac{\tau_p}{2v_p}v_L^2$
which, along with the maximum acceleration ($a_p$) and the equation:
| $ \tau_p = \displaystyle\frac{ v_p }{ a_p }$ |
when solved for the landing speed ($v_L$) yields:
| $ v_L = \sqrt{2 a_p l }$ |
ID:(14478, 0)