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Perhaps, I'm not sure as 1.3 is a very old patch, you should be on at least 1.4, or SH4 1.5 which is the U-Boat Missions add-on as well (don't call it a patch though, or LukeFF will kill you!)
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I've got the U Boat add on coming this week so i'll stick to shooting things from 90 until it arrives thanks very much for your help
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Hello Nisgeis,
I used that method with Sub Battle simulator because it was the only way to aim torpedoes. The sim did not have a functional tdc, and the manual said to point the periscope and shoot, which only worked for a stationary target. At the time, I assumed that the boat needed to remain stationary, but now , after doing much more work with fire control problems in SHI,2 3 and 4, I believe that your boat could be on the move, as long as your boat and the target stayed on course and did not change speed. I wish I had realized that way back then. Sub Battle Simulator was primitive by today's standards, but the gameplay was the equal of anything we have had since. At any rate, I have not used that method with SH4 due to the fact that we have a functional TDC, which is safe and effective when used as directed. the single most effective element of any fire control solution is target speed, and if you have calculated the target speed accurately you should be able to consistently get hits, no matter what method you use.Theoretically, one could use this "new" method even if your periscope is destroyed if you can get a speed estimate while on the surface, then submerge and wait in ambush, and use the hydorphones to detect the firing bearing of the target. I have not tried it but in theory it should work. Thanks for all your help and work with this, it adds to the knowledge base and gives us additional options. Joe S |
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Isn't it weird that the Internationals actually succeeded in defining a preexisting measurement with one derived improperly later!? This is a farce, but what do you expect from people who can't properly measure the earth? The nautical mile is at most a little under 1% off at certain places on earth. Metric measurements are based on non-human parameters (1/10 millionth of the distance between the equator and the pole, 100th the distance between freezing and boiling water, other alien measurements), and as such are not suited to measuring things that humans use, or even humans themselves. Imperial measurements are based on useful things to do, or on a rough relationship to dimensions of the human body. This makes them much more appropriate for human use. The only thing the metric system has going for it is its decimal nature. But it is intrinsically alien and possibly evil.:doh: THAT's what Europe needs: an exorcist!:rotfl: |
Stumbled across this thread just now and thought it might merit a bump after RFB 1.52
After four patrols in early 1943 I have come to develop my own attack technique by trial and error. Maybe it is of interest here (being a constant bearing technique at 50-60 degrees) One of the best ways to prevent duds is by having the torpedoes hit the target at an angle around 60-70 degrees, so the Dick 'O Kane technique results in a lot of duds for me with the perfect 90 degree shots. the biggest problem with getting a perfect solution at removed angles is that your boat moves during the setup. thus, a perfect solution at 1500 yards on a bearing of 330 will have turned into an unaccurate shot by the time the target gets there (as observed in the original dick 'o kane video I believe, when the boat drifted a bit after cutting the engines) keeping in mind that realistic gameplay calls for continued propulsion to maintain depth, this becomes undesirable To solve this problem, you can aim the boat so that the angle on the bow will be 60 degrees starboard at your zero bearing. At a target speed around 10 knots, this results in a shot at several degrees starboard angle off your bow, which is still good enough for me. The greatest advantage being that your solution will stay accurate even if you get the range wrong by several hundred yards by slowly creeping up on the target's course line after inputting the solution still with me? It's basically still the same constant bearing setup, for a shot at zero bearing (with the torpedo making a tiny turn to starboard) while the entire boat is not on a nice 90 degrees to the target but rather a dud-preventing 60 degrees to the target. Personally, I find this an excellent way to get some tonnage in the log, even with the early war duds. If this is all one big unclear mess, I'll provide some screenies :) |
Works perfectly for me. If you use the vector analysis from the video, which you can do in-game right on the chart, you can calculate the correct lead angle for any angle to the track. Then you can throw the TDC overboard (saves weight for more food!) and go to town!
It does mean that you have to know the torpedo's speed, but that's a minor difficulty compared to the flexibility you get. |
ooohhh just watched the vid (nice one btw, liked the taskswitching and drunken sailor bits :p )
the vector analysis is a new one for me, i'll give it a try on my next contact but what i was trying to say is this: if you set up the shot on a bearing of 349 at point A, for an aob of 34 it won't be aob 34 on bearing 349 at point B, 1000 yards ahead of point A. right? argh all this algebra makes my head hurt. i'll play the game and draw some lines, then i'll probably end up coming back here and saying something like 'excuse my utter stupidity' meanwhile my above mentioned targeting at 0 bearing on 60 degree shots is very satisfying though :) edit: erm although the vector analysis would suggest that the angles will stay exactly the same while the range decreases, as they all converge at the same point of BOOM |
oh *insert random profanity*
You are obviously correct i made a very silly mistake: the dot on the map where the target course intersects with the 349 mark on the bearing ring DOES change as you creep closer to the course... but so does your distance to the target and thus, the angles stay the same, whether you are at 2000 or 500 yards out. as such, range never even enters into the equasing, except for drawing the plot and putting a random number into the tdc to set the bearing in fact, shooting at zero bearing means the torpedo has to make a tighter turn at closer ranges (because the turn starts after the torpedo goes underway, leaving less room to properly make the turn) so if anything, it is the zero bearing shot which will get less accurate as you drift closer firing by vectored lead angle is a gem! I always just took a wild guess, being satisfied with anything between 350 and 10. But now I noticed my attackplot showing a perfect straight line for the torpedo, along the the zero bearing line. torpedo gyro angle is exactly 0 at any range. if you can forgive a poor fool for opening his mouth, I'll just move on and sink some ships. After this load of nonsense my addition to this topic can be summed up with: '60 degrees is also nice' |
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So why doesn't the vector analysis show this? You're going to need a drink. Maybe several drinks.... The vector analysis is done from the point of view of the torpedo, not your sub. The angles all stay the same for the torpedo from the moment you fire to the moment of BOOM. That's because it is supposed to be on a collision course!:up: |
before i wander off to look for that drink...
http://img396.imageshack.us/img396/2860/angleskn6.jpg http://img396.imageshack.us/img396/a...pg/1/w1024.png i imagined a target with course 60 speed 12, the vector in the top right corner shows me that for a 46 knot torpede, the lead angle is exactly 12 degrees so i made a plot at around 1100 yards, moved forward a bit and made a plot at 700 yards. what i tried to show is that the point where the target is when you launch the shot varies between both plots, which is why i thought that drifting away would make the solution inaccurate. (which is why i always shoot at a zero bearing plot, resulting in a 12 degree gyro angle for the fish) but doing the plots in the below screenshot, i found that the angles stay exactly the same. obviously, because the target is closer to the impact point when you launch at closer range, but the torpedo has to travel less far as well, http://img89.imageshack.us/img89/1885/angles2ub6.jpg http://img89.imageshack.us/img89/ang...pg/1/w1024.png so yes the colision stays the same, regardless of range. my mistake (i think we're both saying the same here, but i wanted to show what i learned just now) edit: yes we are saying the same. '4600 yards against 1200 yards' amounts to the same as '46 yards against 12 yards', so the vectors (point of view from the torpedo) never change indeed im gonna get that drink now and use that vector thing for real |
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Aaronblood is laughing at us. Numbers are like cotton candy to him. I think he BREATHES numbers in his sleep. At will, he can make me look like the idiot I am. I'm just a communicator, not a numbers guy. I can eventually make sense out of them, but numbers are aaronblood's native language.:up: |
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http://education.yahoo.com/homework_...266-5-pr-q.gif You can see that the triangles QRP and UVP are similar. It doesn't matter if the target is at Q or U, if it is travelling along the line to R or V, the torpedo will connect. The target could be anywhere along the line QP (or beyond Q) and the torpedo in this picture will connect as long as the torpedo is launched at P toward R. Note that the targets at Q and U are traveling at the same speed and same direction. Also note that this picture doesn't describe what the sub itself is doing (in terms of speed or direction). This picture just describes the torpedo and target tracks. |
well, quite :know:
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Starbird, I believe that's what would be called a classically elegant illustration.:up:
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