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SUBSIM: The Web's #1 resource for all submarine & naval simulations since 1997 |
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#1 | |
Hellas
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i will send you the prove of this at your pms. if not...then it is a fault at your map drawings or caused becuased of inaccuracies of silent hunter. send me the mission you played to give it a try and i will tell you my results
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Knowledge is the only thing that nobody can ever take from you... ![]() Mediafire page:http://www.mediafire.com/folder/da50.../Makman94_Mods |
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#2 | |
Watch
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The need for changing course, either for the fourth bearing or for any other depends only on the circumstances. The fourth bearing won't be in general equal or parallel to the red line even though you don't change course. Actually, in my explanation of the method, you could draw a straight line in the diagrams representing submarine's course and the solution holds, because it does not depend even on the location of the submarine. But in certain circumstances, such as those in Makman's video, you may need to change course in order to obtain a non parallel bearing and increase accuracy of the solution. I insist: is circumstantial. Something that catches my attention is that, if I understand well, Karamazov is not taking a fifth bearing to calculate course and speed of target. Four bearings is enough to know that, but it implies repeating the calculations backwards through the bearings and finding an initial red line through which to find the initial position of target via first bearing. Is that what you are doing, Karamazov? Are you maybe estimating true course by sight from those used in the construction? Are you doing anything else I can't think of? Edit: sorry Karamazov, your picture made things really clear for me. You take a fifth bearing. Silly me. Thanks for your comments. |
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#3 | |
Hellas
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i insist that if you DON'T CHANGE COURSE AND SPEED the fourth bearing will be ALWAYS the red line ! all these only at the situation that you haven't change COURSE AND SPEED you say: ''Actually, in my explanation of the method, you could draw a straight line in the diagrams representing submarine's course and the solution holds...'' this point is not correct,imo, Kuikueg. if you draw a straight line for your course then the spaces between the time intervals will not be equal (which can be 'translated' as a 'speed' change)
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Knowledge is the only thing that nobody can ever take from you... ![]() Mediafire page:http://www.mediafire.com/folder/da50.../Makman94_Mods Last edited by makman94; 07-25-10 at 06:52 AM. |
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#4 |
Watch
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Makman is right!!!
Not absolutely, but right. If you keep speed and course through all the process this is what happens: The construction of a possible solution course (among the infinite existent) implies choosing a point along the second bearing. What if you choose your own second position as a starting point? you will end up with your own fourth position as the protracted one for your target. Therefore, the red line always goes through your fourth position. So if you don't change speed or course along the process, your fourth bearing will be meaningless. In the document that explains the method, I had considered meaningless the submarine position, so I put on some arbitrary ones and all went well, because it seems that slight variations on speed or course result in wide variations in the red line, under feasible circumstances, which is enough to provide an accurate solution. I was quite perplex when I realized that, after Karamazov's statement that he had not changed course. After reviewing his snapshot, I notice that he had at least changed speed between his first and third bearing. That's enough. I have yet to assess to what amount the direction vector of the red line is sensible to those changes, but I can assure you that you need at least one change in speed or course anywhere along the construction to make your fourth bearing meaningful, (not necessarily right before the fourth bearing, and that's the only reason I said not absolutely right at the beginning, but he is essentially right: it is not that your bearing will coincide with the red line, but that it will cut it through your own position, a point that we mathematicians call a singular point -and we sailors call collision point- where information has vanished. That is, roughly: the bearing to a ship in your own position is any). I will rewrite the document posted in this thread to reflect those facts and post it here. Thank you, Makman. Well seen. This is collaboration at its best. |
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#5 |
The Old Man
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I've played around with that GeoGebra program you sent me, what an amazing tool, I'll always keep this installed. I used my previous bearings for practice. I didn't even need to use my sub's position on the second bearing, somehow the red line always coincides with the fourth bearing. I have to admit, I feel stupid right now. Not because I was wrong before, but because I must've done SOMETHING wrong to end up with good solutions and now, for the love of God, I can't remember WHAT
![]() ![]() ![]() So Makman, you have my apologies for not initially believing you. So.. on the move, after 3 bearings, we can can the 4'th. I still can't shake the feeling that knowing what distance we've traveled between readings, we could somehow protract the bearings back to our original position. That would solve this problem ![]() Wait, I've just thought about a counterexample to what I've just said. Imagine moving along at a constant speed and course and all first 3 bearings are 90 degrees. This means that the 4'th will also be 90, this can either be a ship moving parallel to you at the same speed as you, or at any angle to your course but at a greater speed. Yeap, this problem is unsolvable... Last edited by karamazovnew; 07-25-10 at 12:47 PM. |
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#6 | |
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#7 |
The Old Man
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I did, I meant that the problem of 4 bearings while your sub is holding a steady course and speed is unsolvable.
And I managed to figure out why my solutions were correct. Because of the 0.5 ruler round-up, I ended up with errors in my position which translated into differences in my sub's speed. Since the distances and ranges were very small, this was enough to give me a workable solution. At longer distances (>10km), I would've ended up with the redline overlapping my 4'th bearing. So when I scaled up my plot, making my sub travel 1000 yards instead of 100, the error went down by a factor of 10 and I immediately ended up with all protracted points on the 4'th bearing ![]() |
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#8 |
Nub
![]() Join Date: Nov 2010
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Hi all..
I've downloaded Solution Solver, but a can't open it on my lappy. Anyone can help me? |
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#9 |
Sea Lord
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Pardon me if I may say, not having played SH5 but spent many 'game years' in SH3/4 on 100% (full real - no icons/no charts/nudder/nothing), there's nothing to beat target course and speed by either.
1) running parallel to target course. 2) running 90Degs to target course. with both of these methods, and some experience you need at most 2 readings to determine target course and speed. It's simple mental maths and no chart drawings.. etc. ![]() |
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#10 |
Sea Lord
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Great info, but until point 1. and 2. you need to figure out the target course &speed.
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bearing, tracking |
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