Это почему "великовато" ? Йорктаун по справочнику развивает скорость в 34 узла или 38,5 миль в час . ------------------ Happy Landing !
Меня интересует скорость торпеды и, в конце концов, флотильи в нашей игре. Кто знает? ------------------ Luck Yedyge aka exec[228] aka Killer Crayon mail2duc@stones.com
Torpedo S.O.P. Well, I have attempted many methods, and this one seems to work the best. Note that Down the Throat and Up the Ass shots work well too, but you must be very accurate. A LITTLE BACKGROUND Obviously I have had a lot of experience in shooting torpedoes. It is my life, my job, my forte'. But, today's torpedoes are much more advanced than WWII torpedoes, and the real problem for submarines lies not in firing the torpedo, but finding the target and determining what course, speed, and depth he is on. We do all of this acoustically and cannot sight the target, nor call him out to others. Luckily we do still practice WWII techniques that apply in WarBirds. The whole goal of shooting a torpedo, today and in WWII, is to match the target's speed across the Line of Site (LOS) with the torpedoes speed across the LOS (also know at "match bearing rate"). Speed across the LOS is the speed the target is making along a line that is Normal (90 degrees) to the bearing that you see (or hear) him on. (see pic below) A targets Angle On the Bow (AOB) is the angle from his bow to the LOS (the bearing to him you are looking down). If you are looking bow on, AOB is zero. If you are looking from directly astern, AOB is 180. When you are abeam of a target, AOB is Port (or Starboard) 90 degrees. To find the targets speed across the LOS, you multiply his actual speed by the Sine of his AOB. if V=target speed and Vx=target speed across the line of sight, then Vx = V * Sin(AOB) We want the torpedoes speed across the LOS, V(t)x to be equal to the ship's speed across the LOS, V(s)x, so that they collide. And since the torpedo is travelling faster than the ship, it can lead the target enough to match Vx and still be closing him fast. Every time you shoot an arrow, or a bullet, or pass a football you match speed across the LOS and hit the target. Same thing here. I spent some time attempting to get relative speeds of the ship to the torpedo. There are no absolute values I can get, but I found that the torpedoes travel about twice as fast as the ships. Close enough for government work Now we know that speed of the torpedo is 2 * speed of the ship, which mathmatically is V(t)=2V(s) where t=torp and s=ship We set V(t)x = V(s)x (which means speed across the LOS for torpedo and ship are equal) We drive the plane to be directly abeam the ship when dropping the torpedo, so we know his speed across the LOS is his actual speed. V(s)x=V(s) This is true because his AOB is 90 and Sin90 = 1 Now we just need to shoot the torpedo at an angle so that the torpedoes speed across the LOS, V(t)x, is equal to the ship's speed. If a ship were to look at a torpedo incoming, he could define an AOB for it. From the shooter, we call that angle the Lead Angle (L), to avoid confusion. Just as you would lead a receiver with a football, you lead the ship with the torpedo shot on a lead angle. So we need to shoot the torpedo on an angle so that the torpedoes Speed across the LOS, V(t)x is 1/2 it's actual through water speed, and hence equal to the ships speed. They will then collide, and you will get a hit. Solving all the above equations you can find that the angle required is 30 degrees. V(t)x = V(t)(sin(L)) = V(s)x = V(s)x(sin(AOB)) Assume V(t) = 2V(s) Drive plane to get sin(AOB) = sin(90) = 1, and then V(s)x = V(s) 2V(s)sin(L) = V(s) divide by 2V(s) sin(L) = V(s)/2V(s) = 1/2 and the sin30 = .5 = 1/2 L = 30 In real life the plane is moving, and turning, and the torpedo drops at a speed (horizontally) much faster than it goes through the water, and the torpedo must start up and get up to speed once in the water, all complicating the problem. I have found that while turning, dropping at 30 and 35 degrees from your starting course (with the ship directly on the bow) will get the torp a hit if you do it just before the ack starts shooting at you. And all this was done with a JU-88 and 2 torps onbd. Should work with any plane though. Summarized, the strategy goes like this 1.Drive the plane on a similar course as the target, but offset so that you will pass him well outside ack range (twice as far) directly abeam him. On this leg go to 70% rpm and use air brake to drop to < 200kts, then secure the air brake. also drop altitude to < 200ft. Ideally I tried for 175 ft at 180 kts on same course as target, directly on his beam. Make sure to turn off the air brake or next turn will give you a face full. Also, 70% is what I found maintains 180kts once you are on speed. 2.When you pass through his beam, turn hard into him being careful not to auger, and place his centerline or bow (nothing astern of his stacks) in the middle of your crosshairs and drive in. Maintain 180kts at 175ft (not always easy). Keep him in the crosshairs, trim level here for the next leg. Drive the plane in to just before the ack start to fire (based on you experience), and note your course. Now turn through the targets bow and while passing through 30 and 35 degrees from that course, drop your torps. Note you must maintain <200ft and <200 kts, while not hitting the water , turning, watching course, and dropping torps, etc. 3.Now throttle up and get out of there. If you time it right, you will never get any ack fire. Obviously this takes practice. and lack of bogies, or fighter escort is probably required. Let me know how it goes. lt-d ______END_______ Вот, когда-то где-то в инете нарыл! CYA
Кстати Айке вчера сказал, что теперь авианосец полностью восстанавливается от повреждений после трех поворотов. Я вот только не понял, а если 1 бомба до 1-го, 2-ая бомба после 2-го поворота, то как тогда? ------------------ See you in sky мл. лейтенант 228 ShAD VVS RKKA
Если не трудно, скинь данные по торпедам, которые использовались в авиации и их сравнительные характеристики (если есть принципиальные различия) с использовавшимися на подводных лодках. Применительно к WB особо интересны японские и немецкие. Если информация слишком велика для форума скинь пожалуйста на serkus@inbox.ru Заранее благодарен. Всем удачи, serkus.
Клева. Жаль, в русскоязычном форуме, да на английском языке. Кста, чувак завершает словами - "Попробуйте да расскажите". Ты пробовал? Вообще говоря, излишняя прямолинейность флота в WB плюс забытые знания тригонометрии дают простой ответ: используйте теорему синусов для решения сего торпедного треугольника! Слабать мелкую прогу для таких расчетов - 5сек. ------------------ Luck Yedyge aka exec[228] aka Killer Crayon mail2duc@stones.com [This message has been edited by -exec- (edited 30 June 2000).]
to serkus: блин с етим компом забыл куда задевал всю инфу, а так нашел не все : USA: MK-13 ,калибр 569,5mm ,заряд 272 кг ,общ вес ~ 1000кг, скорость 33,5 узла, дальность 5210м; МК-13 выс сброса до 37м скорость 241 кмч,торпеда МК 13.3 могла быть сброшена с высоты до 305м при скорости 555 кмч. GER: далее для того чтоб пальцы не отсохли так : -тип торпеды/калибр мм/вес кг/вес ВВ/ скорость хода узлов/ дольность хода м/примечания. -F5B/450/670/240/33/2800/ выс сброса 10-43 м при глубине моря не менее 20-25м -F5B модифицированная /450/812/200/24 узла дальн хода 6000м или 40уз -3000м время кончается продолжение следует Vik
Все просто. CV восстанавливается полностью в каждом третьем повороте. Вчера его опустили до 4%, потом у него опять 100% и заново.....
Учитывая легкость Ки-43, я не пробовал только усадить его поперек палубы. Самолеты потяжелее - проблематично. К примеру, меня аэрофинишер останавливал как раз за кормой.
