FAR § 25.493 Equation 1

[SteveAero]

Hi,

FAR § 25.493 (e) states an equation to calculate the maximum nose gear reaction in a braked roll. I cannot for the life of me back-engineer this equation-and it seems so simple. If anybody could show me how to calculate it, I would be most appreciative.

Here is the relevant excerpt and Figure 6 from Appendix A is attached below.

FAR § 25.493 (e) In the absence of a more rational analysis, the nose gear vertical reaction prescribed in subparagraph (d) of this paragraph must be calculated in accordance with the following formula:

Where:

VN = Nose gear vertical reaction

WT = Design take-off weight

A = Horizontal distance between the c.g. of the aeroplane and the nose wheel.

B = Horizontal distance between the c.g. of the aeroplane and the line joining the centres of the main wheels.

E = Vertical height of the c.g. of the aeroplane above the ground in the 1·0 g static condition.

µ = Coefficient of friction of 0·8.

f = Dynamic response factor; 2·0 is to be used unless a lower factor is substantiated.

 

[Sparweb]

You had a pretty rough first year at engineering school then...

That formula is a plain static balance of forces equation. Rearranged to solve for the nosewheel vertical reaction force.

I suggest you draw the plane as a free-body diagram and solve for moments about the main wheel axis.

 

[SteveAero]

Oh @Sparweb, master engineer, please forgive this dumb question from a lowly underling, and please also provide a worked solution-so I can finally pass 1st year mechanics.

 

[berkshire]

So, this is a student post then?
B.R.

You are judged not by what you know, but by what you can do.

 

[SteveAero]

No berkshire, I was being facetious. I really am struggling to rework this equation and could use some professional help. Seems like all I'm getting is criticism...

 

[3DDave]

Moderately interesting - my algebra isn't so bad; the only factor that stands out is the dynamic factor "f" which likely accounts for bumps and such.

Solve for moments at the nose wheel contact with the runway, noting that the force on the main gear is from both the nominal weight on wheels but also is relieved by the overturning moment from slowing the plane that shifts some of that force onto the nose gear. Essentially solve the horizontal acceleration in terms of the normal force on the main gear.

Once one determines the vertical force available then from friction that is fed back to determine the horizontal acceleration.

Knowing the horizontal acceleration, one takes moments around the main gear - the moment the normal force the nose wheel reacts to. Keep track that each part of the expressions have the correct units.

 

[SteveAero]

3DDave-I tried that. I still do not know where the μE term comes from in the denominator. Can you post your worked solution?

 

[3DDave]

I could re-do it, but why? If you need the equation - there it is. If you have need for algebra help, post your work so far.

 

[Ng2020]

Here you go. A bit clumsy.
I'm sure there's a more efficient way to the same solution, but it works.
Shown nearly every step for clarity
The f is tacked on after the fact.


Note for simplicity V_M is the total main gear reaction in my freebody. (Whereas faa is 2V_M)

 

[SteveAero]

Ng2020, you are a genius!

I see where I went wrong:
1. I assumed from Figure 6 that there is a friction force also applied at the NLG-which I see only now applies if the NLG has brakes. Do you know perhaps if the A330-300 has brakes at the NLG?
2. I worked through the problem with the dynamic factor included - not applying it at the end.

A thousand thank-yous. You are a truly helpful engineer.

 

[Ng2020]

No problems, glad it helped.

I'd say with reasonable certainty the a330 does not have Nlg brakes, they're quite an uncommon design feature. Might be available as an option, but unlikely.