ADMIRAL_PENKOVER

Nápad na stejným způsobem fungující, ale jinak tématicky laděnou akci: "Studená válka" Team 1: SSSR Team 2: Všechno ostatní. Samozdřejmě team 1 získává více bodíků, aby dorovnal početní nevyrovnanost.

Mě Oktyabrskaya Revolutsiya, což by sedělo, jelikož ji už mám a jsem s ní spokojen. :D

Jen jsem se nejak nedocetl, mozna jsem to jen prehledl, co se vlastne stane s letadlovou lodi na liche urovni, kterou hrac vlastni? Napr mam ted Rangera, az se to stane, tak se mi u nej jen zmeni cisilka (jako treba uroven) nebo ho ztratim a budu si muset koupit indepedence, ktera uz vlastne bude jmenovat Ranger..... A co kdyz nekdo indepedence prave ma ? Jako nejjednodussi moznost pro sebe vidim co nejrycheji se vyexpit na lexington, asi ale spotrebuji vsechna volna XP a hromadu signalu. Kdy ze to presne spusti? A co je dulezite, jestli ted Lexington koupim, neztratim ho pak a nevratim se na na langleye, ze ne?

*edited* BTW, at least 3 votes would help me a lot, but i did what i was meant to do.

Thanks you all, now it works fine and I am going to always hit now. Problem is when the ship is not pararell or its maneurating.

Are the physics in WOWs same like on earth? My numbers are from wyoming. problem 1) If this strange numbers when in sniper mode are degrees, so: I was trying to estimate the position of some ship, where it will be when my shots come there? Distance: about 11km Shell flying time : about 6 s. Enemys ship speed : 30kn = 15 m/s So after 6s the ship will be 90m away from its current position. The ship is about 150 m long soi i already know, where to shot. But i want to use theese strange numbers on the indicator. I have a formula : b=a*sqrt(2*(1-cos©)) where c is angle what we are searching between its future and current position, a is distance from me to enemy and b is distance where enemy will be when shots come. c=arccos((2a2-b2)/2a2) a=11km, b= 0,09km so c=0°28´ ehm, what ?! half of a degree? But the ship is 150m long and after 6s will be 90m away, on my indicator it is about 5°. When i try to fire on that 5°, i hit. Am I counting wrong ? problem2) Formula about shell flying time: if we use numbers from problem1 :a= elevation angle to achieve 11km. g=10m/s, t is flying time of the shell (we expect 6s), v0 is 884m/s The elevation can be counted from this formula: from where we could see that the angle is 0,4° (more about this see problem 3) so the flying time is :1,22 s. but it shows me 6 s (see problem 1)...........I am going to make a suicade. problem 3) Shell start firing speed is 884m/s g=10m/s we have another formula : where v0 is starting speed of the shell and a is elevation angle of the gun. This should show us the max firing distance. a= is somewhere in <20°;45°>. If we say a=45°, the max firing distance is 78km! If we choose 20°, the distance is 50km. Are we doing something wrong? I dont count the aerodynamics, but, i hope, it wont decrease range by 60km. If we go to::: https://en.wikipedia.org/wiki/16%22/50_caliber_Mark_7_gun :::we can see, the firing speed of Iowa class BB is 762m/s for AP. Gun elevation is 45°. It means 58km - aerodynamic, what could be somewhere about 30km. Reality is ok. ...!!!!!!!... PS: the screenshot is from google.