I personally started playing war thunder because it was the most realistic free game that is out there and was striving to become more and more realistic as it has ingame. (I started playing 2013).
While this is one of the possibilities, there is another way this can be done. Most Radar scans carry IFF interrogation with it. If it checks, transponder responds. This can also be used with RWR, as when radar ping carries IFF that transponder responds to, it can be marked by the system as friendly.
I don’t like the map spawn cancel change. There are often times where I go to spawn in, and then realize A or C side needs help. Trying to move my mouse all the way up to the tiny cancel button on a 4k screen is impossible to do in time.
Why not return the funciontality with a double click on the map?
Better yet, allow us to click on A, B, or C and it will automatically determine the fastest spawn to that objective (taking into account hills, roads, trenches).
There are some people who used to play a lot of Sim games, who simply play WT because the game is more fast paced and isn’t that demanding in terms of playing, since you got instructor, mouse aim and third person view.
Realism in WT is somewhat far fetched. The damage model is very simplistic, certain ammo types are often severly lacking and of course you can pull way more Gs that should be possible.
Wounding a pilot also has no effect on the performance anymore.
Realistic battles is practically the arcade equivalent of a real Sim and the biggest difference between Arcade and Realistic, other than the lead-indicator, is honestly that you can spawn more planes and battles are even more action oriented, since you don’t need to climb for 5min before you run into an enemy.
Annoying F-4S and F-14As got their biggest weakness buffed. Another buff for the US mains. I was able to play the red team because Viggens and Mirages could easily kill them with radar missiles because they have all aspect PD radar and AN/ALR-45(V) cant detect PD signals. Fishbeds and Floggers can now also detect their radar but with shitty SPO-10 and FM they are already gonna lose against them anyway.
Also it would be better if you just made us able to toggle RWR IFF and not just delete them completely. Maybe i would be able to defend myself from brainless teammates just locking and slinging a FOX-1 at me.
And i am playing “Simulator” a game mode thats supposed to be more “realistic” than RB. If an aircraft have weaknesses it shouldnt be buffed with unrealistic features, theres another option called “Battle Rating”. I play both the blue and red team but now it seems like only blue team can be played because theres always more blue players than reds and reds cant win like that. I dont want both AN/ALR-45(V) and SPO-10 to act unrealistic. I have seen a lot of players fly aircrafts with SPO-10 and still do very good against F-4S and F-14A
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.
very bad drawing to explain
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.