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#1 | |
Sparky
![]() Join Date: Apr 2013
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I know that A^2 + B^2 = C^2 in a right triangle from geometry. Now that works for right triangles, those with a 90deg angle, which just so happens to be exactly the angle you want to attack your torpedos with on the hull of a target. So I say screw trig. I dont have time to do trig in the boat while lining up the ship, and keeping the boat in order, checking out asses for destroyers making sure there are no AC etc. Even the best captains under that level of stress and daily grind, do you think any of them would want to put their trust in complex mathematics when the KISS keep it simple stupid principal applies? Bologne, they want to sink the the ship and go home. Just think one complex calculation with a single error could throw off the torpedos. Why use complicated stuff when you could use easier stuff. Of course if they had access to GPS they'd probably use it too but bah they didn't so why would we in a "simulation" LOL turn on GPS cheating. Come on guys. What kind of Skipper are you, are you on a 3 hour tour with Gilligan? Simple geometry easier faster to calculate some in your head even. A+B=C My favorite right triangle is 3^2+4^2=5^2 9 + 16 = 25 |
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#2 |
Henke
Join Date: Feb 2012
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I always liked
A^3+B^3=C^3 ![]()
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#3 |
Grey Wolf
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Well I was finally able to try this. I discovered to my chagrin that the line of bearings I made did not intersect at all. Instead they fanned out. This gave no useful data at all other than the contact was moving. I then turned contacts back on, and tried this again, and saw that the termination point of the lines of bearing did kinda follow the course of the contact, but were many km off. Still need help here.
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#4 |
Sparky
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Can you show or tell us list us your observations...
Sometimes the lines of bearing do fan rather than converge but this indicates the target is move faster thank you or the course you are using against their course is too wide or even perpendicular to their own. Lets debug it. |
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#5 |
Sparky
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#6 |
Grey Wolf
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It was a destroyer, and what you said is what I was thinking, the contact was going fast. So does that mean this method won't work on fast moving contacts? I got the initial report, set my heading toward the contact (smoke on the horizon), set my speed to 10 kts, marked my point, and drew a 30km line. Waited 2 minutes, and placed the next mark, I forget the bearings given. As I stated though, the lines fanned out, and the terminiation point of the lines if a line was drawn from point to point would almost parallel the course of the contact. (recall I did have map contacts on).
Hope this helps. |
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#7 |
Sparky
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Im having trouble visualizing what you were observing. There are times when triangulating is going to be difficult or impossible. If the contact is moving away at high speed. A good thing for a destroyer, you would typically see a converging of those line of bearing farther and farther away.
For example if you drew a circle every time around the location where your last 3 lines of bearings get closest to each other and then each time you did that those circles would move gradually would be farther and farther away even though you should observe the lines of bearing being closer and closer together. If they are starting to fan out and the tighten up that would to me indicate a destroyer running perpendicular. One running parallel to your own subs course. Would be relatively easy to spot so long as you are both heading in the same direction, if you are heading in opposite directions the angles between the lines of bearing will get smaller and smaller. Maybe we are making this too complicated. If you have a visual contact to make the general direction of travel as simple as possible observe the smoke from the ship. Think of the ship as a candle, as you move a candle from left to right the smoke is highest in the left where it originated from and lowest at the point closes to the flame. If that smoke is on a steady rise without a trail either on the left side or the right side, that ship is either coming towards you or heading away or stationary. If it is larger in 2 minutes its coming at you. Let me make a few more diagrams to aid you and then we'll see if we can figure out how to determine the course of a fast moving target. If its pressing at 30 kts or so you may not be able to exactly nail down its course without trig but with practice you should be able to guess it within 10-20 degrees based on the sight picture of the angle on the bow and your ship's current heading. Meaning if you are heading North and you see a car traveling from left to right which direction is it heading? It is heading East. Now suppose you are parked at 330 and when looking at the AOB or angle on bow it appears to be at 60 degrees of the right side, so you can fully see the bow and not see the stern, so take a guess at which angle you are seeing. If you are at 330 and the contact is heading left to right and you appear to be at AOB 60 Right then it must be heading about 030. It is a lot of guessing and you wont always be right. Which is why you make more than one observation. The closer you get the easier it is to make accurate guesses, and shorter the torpedo run has to be. As long as you are in deep water there is no benefit to attacking beyond 900-1000 meters and if you can get inside between 400-800 meters. Do it. One torpedo shot from the hip at that range at high speed almost always hits with enough time to launch a 2nd one on a modified gyro angle if the first one fails to detonate. Last edited by Jaystew; 06-18-14 at 07:01 PM. |
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#8 |
Samurai Navy
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Maybe a different example? Smoke in SH5 is effected by the current wind condtitions, therefore Smoke is an indication of wind direction and intensity but not ship course.
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#9 |
Sparky
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Lol what weather? I typically have 0kts for wind speed. So you can't determine if a ship is heading from left to right or right to left because of the weather on your configuration? Hmmm interesting.
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#10 | |
Henke
Join Date: Feb 2012
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