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Gear compression

J

johnor999

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87
Country
sweden
Hello,

I've been searching around and I can't seem to find anything on how FS calculates gear compression.

An example on what I want to calculate:

A given mass m is applied on a landing gear contact point. For simplicity we assume that everything is at rest. (the plane is just standing on a runway) How much in terms of length would the gear be compressed? I can't seem to figure it out.

I'm not making an aircraft or tweaking anything, I just want knowledge.
 
Thank you Roy! My searching skills are non-existant. I do try, and not just 15 minutes but hours/days, however I still fail to find what I'm looking for.
 
Hm.. is the static compression the ammount of compression at empty weight? Regarding the spring constant equation k = F_x/(x - x_0); is x = the length/height of the contact point - static compression; x_0 = the length/height of the contact point?
 
When in doubt, consult the SDK. You will probably get some kind of answer:
"Static Compression This is the distance a landing gear is compressed when the empty aircraft is at rest on the ground (feet). This term defines the "strength" of the strut, where a smaller number will increase the "stiffness" of the strut."

The way it is set up, you do not really need to do any calculations other than those given in the wiki.
Roy
 
Yes I've read that, however if I was to increase the weight, what would the new compression be? Is it just a matter of linear interpolation or is there anything else I need to do?
 
It would depend on where the weight was applied. You would need to see what the rotational moments were at each wheel position according to CG changes resulting from the increased weight compared to the empty weight situation.
If the cg moved back so it was behind the main wheels (assuming a tricycle gear) then the nose gear would extend fully as the beast tipped up on its tail. Silly example, but there were airplanes where this could happen (DC-6 was one I think)
If you know the increased load on the specific gear, the static compression would increase and indeed in a well made model in the sim you can see that happening when, for example, bombs are loaded on it.
I guess it would be a linear interpolation unless the force was sufficient to bottom out the strut.
Roy
 
Okay, I've finally understood exactly how to calculate the compression for a given weight after some testing in the sim. Just in case someone else needs to know this I'll share my findings.

Conditions: Airplane at rest, the weight evenly applied on all contact points, engines off, winds calm.

compression = static_compression * (weight - empty_weight)/empty_weight

With a high enough weight, FS would try to calculate a compression that's longer than the vertical position of the contact point, which results in an endless bouncing due to the fact that the VMO would be below the surface if it didn't bounce up.
Also note that moments would need to be accounted for, but once you know the corrected weight at each landing gear, the calculation above would apply.
 
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johnor999 said:
With a high enough weight, FS would try to calculate a compression that's longer than the vertical position of the contact point, which results in an endless bouncing due to the VMO would be below the surface if it didn't bounce up.
Click to expand...

Not sure I understand you there. VMO?
 
hairyspin said:
Not sure I understand you there. VMO?
Click to expand...
The Visual Model Origin. Every coordinate is relative to VMO. So if I have a contact point at a vertical position of -10 feet, then the VMO is 10 feet above this contact point. Let's assume a landing gear has a contact point at -10 feet and is compressed by 11 feet, then the contact point would be located at -10 + 11 = 1 feet above the VMO. On ground, the distance from VMO to the surface is the same as the vertical position of the contact point. so if the contact point is moved from -10 to +1 the contact point would be located at 1 feet above the VMO. If the surface was 1 feet above the VMO, the VMO would be 1 feet below the surface, which FS doesn't allow.
 
Every coordinate is relative to the reference_datum_position... which is usually incorrectly set to the VMO.
 
RDP and VMO are one and the same, coordinate 0,0,0. What is incorrectly set by newbies at times as the COG.
 
How odd... every aircraft I've created for trainers has a reference datum position (manufacturer documented) forward of the nose with the exception of one that had it at the forward firewall behind the engine. The VMO is not in any of those locations.

For the Citation XLS+ (as example) the reference datum is 306.59 inches forward of the LEMAC.

Imagine how oddly the aircraft would fly if one used an incorrect reference datum but a correct empty CG arm.
 
Ed is quite correct. RDP is (in real life) a completely arbitrary position that the designers create, and from which all subsequent positional datum are derived. Many modelers who either do not understand this, or are just lazy set the RDP to be congruent with the VMO.

The advantage of setting the RDP forward (or aft) is that all positions fore and aft are the same sign; all negative or all positive! Moreover, if one is blessed with having accurate plans from the designers, the measurements can be more precise. :teacher:
 
n4gix said:
RDP is (in real life) a completely arbitrary position that the designers create, and from which all subsequent positional datum are derived.
Click to expand...

Quite so: an arbitrary position which, with the VMO, gives the sim a handle on where things are on a visual model. With the proviso noted by Johan above, the modeller is free to place this wherever suits: if manufacturer's data is available then adopting the manufacturer's position is only sensible; if not then it's the modeller's decision. I note the SDK states if no RDP is specified, a position of 0,0,0 is assumed by default.

I'll be bold and say there's no correct position in any case, so long as everything adds up (or subtracts) accurately. But as the song goes, make it easy on yourself...

BTW, I hope Johan has a better idea of gear compression now.
 
Last edited:
n4gix said:
Ed is quite correct. RDP is (in real life) a completely arbitrary position that the designers create, and from which all subsequent positional datum are derived. Many modelers who either do not understand this, or are just lazy set the RDP to be congruent with the VMO.

The advantage of setting the RDP forward (or aft) is that all positions fore and aft are the same sign; all negative or all positive! Moreover, if one is blessed with having accurate plans from the designers, the measurements can be more precise. :teacher:
Click to expand...
I stand corrected. I had forgotten about being able to set the RDP within the CFG. Lacking detailed information/engineering drawings (LAZY! :yikes:) making measurements within the CAD (as Tom said, makes life easier) when VMO and RDP are coincident. After all it is arbitrary. ;)
 
Is there any influence of "wheel radius" (contact points) to compression?

Is there any purpose for "wheel radius" in contact points.
(FS2004 SDK says nothing about it)
 
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