Do people really think the killcam’s biggest use was revenge CAS? Any competent player in this game uses it to mark the map during that brief 2 second window before spawn screen.
With this change I not only have a difficult time calculating the exact position of the enemy tank, but I also struggle marking the map in time because the killcam is so short.
For those freaking out about CAS, there are a billion ways to make it better, and this change is not one of them.
I quite liked the old one, because knowing where you were shot from causes you to learn new things you can and can’t do.
Finally R-27ER full loadout 550 SP
AIM-120 685 SP
I can finally get my insert plane name and won’t see 100 AMRAAMs
Its not like there is the same problem on EVERY map and count trees also.
Imagine that you can make a update without killing/redzoning/corridoring/spot removal more maps… just wow… now revert all map chcenges.
Is this bugged?
I only see the enemy vehicle and then my destroyed one without the x-ray penetration.
I still have the antenna move away from limits. It appears that the change locks the adjustment during certain (high g?) maneuvers and based on plane orientation (steep climb, inverted). Now I cannot adjust it when banking and maneuvering.
During gentle maneuvers to stay close to ground level with full real controls and the radar set to the lower limit, it will still climb away from that limit.
When inverted, especially after a zoom climb, as I’m working to readjust the radar position to the horizon, it is locked in place and cannot be adjusted until the plane is in level flight.
I actually prefer the way it was broken before to the way it is broken now. Though ideally, it would just preserve the orientation relative to the plane until I change it.
Now fix Landing gears they should not tear off from an aircraft at all.
how in the world did you come to this conclusion?
Short answer:
Landing gear (LNG) endures much greater forces on a ground (landing, breaking, engine test runs etc) the ever can be induced in air by drag.
This is especially obvious with naval aircraft where for example 30-ton F-14 is ejected by catapult with 3-4g acceleration towing it by nose landing gear (NLG) which then fall off in “400kmh”? That is beyond ridiculous.
LNG is the most durable part of the aircraft attached to primary airframe construction, if you manage somehow separate LNG form the airframe it would not be without severely if not catastrophically damaging the airframe.
Reason why there is a speed limit for LNG, is the LNG doors which are way more fragile and surely can be damaged.
This is main landing gear (MLG) of C680, you can see the trailing link holding the MLG doors… yet iaw aircraft flight manual (AFM) such doors can withstand 270kt. Add a safety margin and you got about 700km before something happens to such doors.
Yet in WT whole LNG rips off like nothing.
This doesn’t really make sense though.
The force that landing gear normally see is either along the normal (90 degrees) of the anchoring point (the weight of the plane) or in a rotational point along the wheels (when accelerated at takeoff or landing).
The force from the wind would in contrast be applied as a rotational momentum in relation to the anchoring point and result in a much greater total force due to the area the force is applied to by the wind.
so at launch from a naval ship the forces convert from normal forces into the wheels rotational ones as the wheel starts spinning. while the air has no such conversion so the forces are all applied as a momentum arm at the base.
to give a somewhat figurative comparison:
Think of the wheels of a car, the ground keeps hitting them at +100km/h at the “base” making them rotate. Now instead put a concrete block on the road at the hight of the center of the wheel and drive into it making it hit the wheel, the wheel would most certainly break even though it sees the same amount of force as the road is exerting at the base. (not a very accurate analogy but i hope it gets my idea across).
You better draw yourself a picture, but I see several problematic points with your argument, first air is not concrete and wheel rotation is irrelevant. So, hitting hard concrete obstacle would rip off the LNG with all the consequences (destroying wing or fuselage structure) but this is not the case we are talking about especially after takeoff. Even when the LNG is dropping it is not instantaneous and in WT planes loose LNG during takeoffs, where drag increases gradually.
But if you want to compare I have calculated Fd for the one MLG of the C680 with formula:
v=111m/s
Ro=1,25kgm3
CL=1.5 (worst case scenario it is probably less)
A=0.5m2 (rougly measured myself)
Where if I did not miscalculate Fd=5800N for one MLG in 400km/h seem a lot but…
If you do a engine run up where iaw regulations AC needs to remain stationary with brakes on and takeoff (TO) power where the PWC306 engines are rated 25kN each, then each MLG would be strained rougly by 25000N and it will hold in place just fine.
So 5800N of drag is really nothing compared to forces 25000N applyed during the engine run up.
absolutely, which is why i was VERY clear about it being a figurative comparison for the sake of understanding and not an actual same situation.
it absolutely is not… the forces exerted at ground gets converted into rotational energy. drag forces do not.
this doesn’t matter that much. the constant force will still eventually reach a point where it exceeds the structural integrity of the hinge and/or pneumatic components.
What you fail to take into account in your calculations are momentum arms.
The engines of that specific plane for instance are placed above center off mass and the wheels below. So not only are there the forces you describe but also a rotational component that counteract those forces (i’m not at all certain about how much this actually contributes, but it is there). So the forces on the wheels wont equal the forces of the engines.
same goes for the drag force, you have to consider the gear strut as a moment arm. So its not just straight forces applied but a rotational force exerted on the connection point of the strut to the plane.
You’ll notice that the formula uses exponential speed here. it doesn’t take much more speed (a quick math in my head gives me double? so 800km/h?) to bring those numbers up to close to the 25kn you mentioned.
Sure, the speed Gaijin uses for braking off of the gear might be WAY to low. But to say that they wont break of at all feels disingenuous.
I will however add:
I’m not very knowledgeable in this field specifically but more in general physics.
I don’t actually know how right or wrong i am about this. I just felt like the use of:
felt WAY to exaugurated.
That is a whole point, it won’t (in level flight). On a ground all engines power trasfers to LNG in the air engines need to overcome drag of the airframe as well. So you are unable to accelerate to speed where the NLG might tear off.
BTW that hinge would withstand hard landing of fully loaded aircraft so most likely the “pneumatic system” if you mean tire. Which is quite a few Gs even with dampeners.
I did but left it out for simplification. Since it won’t have great impact on a result. And you equations are missing length of the arm between MLG and NLG therefore can’t be correct.
Even if you achieve such speed in dive you think that after 25kN the MLG would collapse ? There may be even greater force during the after landing breaking where I know the AC can do more then 1G de-acceleration on a brakes only.
And there is an safety margin of 2.0 in case on LNG.
You are making your assumptions based on commercial flights that do static takeoffs (which isn’t common nor all planes can do it).
it doesn’t. this is a very famous physics problem. The one about trying to start an aircraft on a treadmill. the wheels have almost nothing to do with the acceleration on ground unless they are breaking.
to maintain speed the drag needs to equal the force of the engines. To accelerate the engines needs to exceed the drag.
the acceleration has nothing to do with the force exerted on landing gear in the air other than as a function over time. The force in air comes solely from drag as a static pressure. the engine power and the airframe drag has nothing to do with how much force the landing gear can take. Only speed as a changing factor, the other factors are constants.
yes, in a right angle, not as a momentum arm. if you have a metal tube standing on your desk and push it down not much is going to happen. but push it from the side and it falls. the forces you describe are happening in a direction it is designed for, the drag force at M1+ isn’t. There is a difference between compression, tensile and shear forces.
it probably has more impact than you think.
You are absolutely correct, my mistake. F2 should be F2 x L2(the mlg-nlg distance)=M2=M1 .
not immediately. there will of course be designed tolerances. but think of Jet fighters speed in comparison; they are doing M1.0 and above.
again, you have the rotational force of the wheels as well. I also HIGLY doubt that the aircraft is decelerating at 1G at landing. that would be quite unconvertable for the passengers.
No I’m talking about engine run up, commercial or fighter jet the procedure is the same.
And that is exactly what you do in engine run up, fire the engines at max power with parking brakes on.
It does, landing ger out => more drag => sooner thrust equalizes drag => plane stops accelerating sooner, this reaching less speed.
Less speed less drag on LNG.
I have pretty good idea from the NLG strut compression during power assurance check.
not with the LNG out…
neglectable
I did not say it is something you would exercise with passengers, I said that the AC can do it.
Not, it isn’t. there is no singe “procedure”, its different from plane to plane.
Most planes usually go to 20% power (idle) or 40-50% power before releasing brakes (rolling start), a few go to 100% power before releasing brakes (static start) but that is uncommon and only done when needed for a short takeoff.
see above.
oh yes, but that has nothing to do with what we are discussing. like at all. you are not going to reach Vmax before the gears break.
do tell, how much of the force goes into it instead of the brakes?
if lng is extended at those speeds. and yes, i think todays jet fighters could reach those speeds with lng out as they can reach M2.0+ with them folded in.
Are you sure? if say at landing the wheels didn’t rotate they would break of almost instantly from the frictional force from the ground or completely grind away if set down slow enough. all that initial energy goes into rotation.
true. But again, i highly doubt it.
V^2=U^2 + 2as
lets say landing speed is 125kt (~64m/s) and the goal is zero speed with a deceleration of 1G.
64^2=0 + 2 x 9,8 x s
(64^2)/(2 x 9).8 = s = ~ 209m
are you saying that the C680 can full stop in 209 m ?
No, I’m not talking about take off procedure. I’m talking ALL the time about ground engine test run. Full power, brakes on.
No you are not going to reach Vmax with gear out.
few hundred kilos
It would what but that does not matter if you are interested in strain on the LNG once you slam the brakes.
It is more like 400 since it is not a constant 1G rather the peak.
you wrote:
so i assumed takeoff. you have not written about engine test…
you and i are using different definitions of Vmax. i meant max possible speed at current conditions, not what the aircraft is capable of at optimal conditions.
so a few thousand N? that is quite a lot in comparison to the 25kN thrust per engine. if its like 500kg then that equates to like 10% of total engine power att full thrust (to tired to do the moment arm conversion).
if you slam the brakes to the point of wheel stoppage then i would agree.
sounds more reasonable. the way you wrote before made it sound like 1G sustained.
Engine run-up is the test…
What is an engine run-up?
Definition- An increase of engine RPM to a high power setting that is for testing an engine or aircraft components and aircraft systems. All high power or extended engine runs must be done at a designated run-up location.
It is accurate enough to demostrate that LNG would not go in 400-600km/h like now. I’m fine if LNGs would snap at Mach speed or so, but those low speeds brake offs are super annoying and unrealistic.
Still even some of the forces “leaks” through wheel rotation, rest is still significant. For a reference, yesterday I pulled logs from IRS and AC did peak -0.4G on a normal landing…
hello! I have a question: plz look at the website from Bilibili. 健康行动 93 (2.37.0.60 更新) - 哔哩哔哩 (bilibili.com)
I find sth really weird as the fig show below.
