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#2 |
Ace of the Deep
![]() Join Date: Jun 2005
Location: Palm Beach, Florida
Posts: 1,243
Downloads: 53
Uploads: 8
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That's a good question.
![]() What you've plotted is a nice DRT TMA track. I haven't had the time to add a MoBo tool that will give a solution like that (even though Dr. Sid has given me the code for it.) I guess I should try to incorporate Dr. Sid's least squares solution tool... ![]() The TMA tool I have in MoBo presently is for working on a maneuvering board assuming you understand the concepts of relative motion. The plotting is actually a little easier (you only need half the nodes you plotted). Here's your DRT tracking data: (true motion) ![]() I'm interested in knowing the relative bearings for each observation: ![]() Now here's how you plot it on a maneuvering board (relative motion): ![]() It looks like 60° is a fair solution for direction of relative motion. I'm going to make a guess at a distance of 9000 yds for the last observation at 1535. You'll notice the diamond appears on the 9th ring of the maneuvering board. I replace the TMA line with a contact and solve for the relative speed of 4.5 knots. ![]() This is where the distance assumption can play into the speed determination. You were nice enough to tell me the speed was 7kts so I was able to backsolve for the 9000-8500 yard distance. If you don't ping for distance, or already have an idea of speed (maybe the sonar man says moving slow) then you would have to plot 2 or 3 more predicted bearings and then shoot off in a different direction for a triangulation. Let's just say, that I was able to expertly determine the distance to target at time 1535 to be about 9000 yds. So, I now have my relative motion solution of 60°, 4.5kts. Now I add the contact RM vector to my ownships vector and... ![]() ...I get a true motion solution of 90° and 7.2 knots. So Captain, target bearing True 1°, Rel 236° at 9000 yds, course 90° at 7.2 knots. Probably a good enough solution to score a hit I think. ![]() |
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