How can I calculate the force applied on nose gear during landing as nose gear touches the runway little after the landing gear? I understand force calculations in this question but I think it applies on main gear. How is the force on the nose gear calculated?

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    $\begingroup$ It depends on the angular speed of (de)rotation, i.e. how much quickly the fuselage rotates from landing attitude to horizontal... $\endgroup$
    – mins
    Jul 24 '16 at 13:43
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    $\begingroup$ I think you want to start with energy then find the force. $\endgroup$ Jul 24 '16 at 17:38
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    $\begingroup$ F = ma, the hard part is figuring out how much mass is on the nose wheel. Ideally at touchdown its not more than a minor percentage of the total aircraft weight, as the goal when landing (small aircraft at least) is to keep the nose wheel up until it comes down gently on its own. $\endgroup$
    – Ron Beyer
    Jul 25 '16 at 1:49
  • $\begingroup$ I calculated the mass on the nose wheel but i am not been able to calculate the acceleration when nose wheel touches down the runway $\endgroup$
    – Umair
    Jul 25 '16 at 4:51

Calculating the loads on the nose and main landing gear isn't really different. The answer you linked is a highly simplified approach you should be able to use for the nose landing gear too, but I will provide a "less simplified" approach to calculate the vertical forces. The braking force is already greatly described there.

A simplified approach

First you need to know the impact speed and the spring and damping characteristic of your landing gear.

At touch down there are three possible cases:

  • Main landing gear touches the ground first. Then the airplane rotates down with an angular speed $\omega_L$. The impact speed of you nose landing gear is therefore $$V_{Impact} = \omega_L \cdot d $$ $d$: distance between the main and the nose landing gear.
  • Nose landing gear touches the ground first. (For now empty, if I get time I'll write about it later)
  • Both landing gears touch the ground at the same time. For a glide angle $\gamma$ the impact speed will be: $$V_{Impact} = V\cdot \sin(\gamma)$$ $V$: airplane speed

The stiffness ($k$) and damping characteristic ($c$) of the landing gear depend mainly on the suspension and the tires used. Tires inflated with a wrong pressure can be really dangerous as they can increase the load on the landing gear.

Putting this data together into a mass spring damper model like in the image

Mass-spring-damper model from Wikipedia

gives you an equation for the nose landing gear movement: $$m\cdot \ddot{x}+c\cdot \dot{x}+k\cdot x = -m\cdot g$$ For the reduced mass $m$ in the model use the mass carried by the nose landing gear. $$m = LM \cdot \frac{d-d_{CG}}{d}$$ $d_{CG}$: distance from the nose landing gear to the center of gravity

$LM$: aircraft landing mass

$g$: gravity constant

The initial condition for the differential equation is: $$\dot{x}(t=0) = -V_{Impact}$$ The impact occurs at $t=0$.

To calculate the force on the nose landing gear you just have tofind the maximum acceleration and add the static and dynamic loads: $$F_{max}=-m\cdot g -m\cdot \ddot{x}_{max}$$

Real world problems

While the calculations above works for really stiff wings, in reality the loads on the landing gear will be (probably) higher. This is due to the inertial forces of the oscillating aircraft structure.

If you don't understand where this forces come from try moving your arms up and down while standing: you will feel small forces on your feet. Now repeat the experiment with two 60t heavy wings as in the A380 and you'll see that the forces are not neglectable. And were do the oscillations come from? The same as with a guitar string: after a small excitation (plucking the strings/ landing impact) the structure (guitar string/ aircraft wings) begins to oscillate.

What other things doesn't this model take into account? Coupling between nose and main landing gear, rotation inertia, nonlinear tire characteristics,...


In order to calculate the force, you should start with an estimation of the angular momentum as the nose sinks and the wheel contacts the ground. That angular momentum am becomes zero as soon as the oleo leg is compressed to its maximum stroke, but that takes a measurable time, from first contact of the front wheel with the ground till the extreme compression of the oleo leg. If you assume that the force f on the oleo leg is constant along the compression stroke, and know the time t it takes, then:

f = am/t

Of course, it's only a rough estimate, but better than nothing...

In order to calculate am, you'll need to know the mass of the plane, its moment of inertia when turning on the main wheels, and the angular velocity of that movement.


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