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07-14-16, 01:24 PM | #1 | |
Grey Wolf
Join Date: Jan 2014
Location: 50.1° N, 14.4° E
Posts: 834
Downloads: 82
Uploads: 5
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Quote:
So we have 3 bearings made with 10 minutes interval between B1 - 71 B2 - 80 B3 - 87 What we get at example there is target course - 22 degrees. (its irrelevant here but just for clarification...) -------- At my cheetsheet equation for spiess line is the second (Get 4 bearing method Spiess line) So this is where we expect our sound contact sometimes at the future. You have 2 options here - to calculate spiess line for moment of the same interval after (10 minutes here), or you can even calculate spiess line for other time, you just need to change the equation. That's actually the reason for 2 equations in my cheetsheet (the first one is more simple, because you dont have to calculate different time interval than you used for getting first 3 bearings) The second, allows you to calculate spiess line after 20 minutes, or 30 minutes here, its up to you how you plan your movement to get their real position. (If you dont know what i am talking about, and you'll get it. - to get real position of target you need to move and cross 4.th bearing from another position with your calculated spiess line). So - From my experience, 10 minutes is too short interval for moving to another position and get real 4th. bearing from there, you dont get far-enough from previous position, so lets calculate spiess line rather with 20 minutes interval. What i mean here is we want to know, where we would hear our contact after 20 minutes from this position, while we will be already somewhere else to hear it from another position and by this situation we will actually create 2d raster to get real position of our target (spiess line is something like our ghost would stay at our previous position and get 4th. bearing from here, while we'll get it from another position and cross of this lines will give us real position of our target). So lets do it! Example to get bearing of spiess line after 20 minutes: legend: B4 - spiess line t12 - time interval between B1 - B2 (first 2 bearings from first example in initial post) - 10 minutes = 10 t13 - time interval between B1 - B3 - 20 minutes = 20 t34 - time interval between B3 - B4(spiess line) - 20 minutes = 20 (yes, this is our decided time interval between B3 and spiess line) t23 - time interval between B2 - B3 - 10 t24 - time interval between B2 - B4 - 10+20=30 t14 - time interval between B1 - B4 - 10+10+20=40 b12 - angle between B1 and B2 - 80-71=9 b13 - angle between B1 and B3 - 87-71=16 real calculation: B4 = B1+arccot( ( t13*t24*cot(b13) - t12*t34*cot(b12) ) / t23*t14) B4 = 71+arccot( ( 20*30*cot(16) - 10*20*cot(9) ) / 10*40) B4 = 71+arccot( ( 2*3*cot(16) - 1*2*cot(9) ) / 1*4) B4 = 71+arccot( ( 6*cot(16) - 2*cot(9) ) / 4) B4 = 71+arccot( ( 6*3.5 - 2*6.3 ) / 4) B4 = 71+arccot( ( 21 - 12.6 ) / 4) B4 = 71+arccot( 2.1) B4 = 71+25.4 B4 = 96.4 So after 20 minutes, we should here our contact from this position at bearing 96-97, we can move now and hear it from another position to get their distance as well... |
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07-15-16, 08:34 AM | #2 | |
Mate
Join Date: Jan 2016
Posts: 58
Downloads: 13
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I still have a few questions if you don't mind: 1) What is distance float? And the "Get spiess line distance" part allows us to calculate how many meters the contact has moved during the measurements? So that dividing by time we get speed? 2) What does the "Target distance float" part accomplish? 3) Are A and B input variables or things which need to be calculated? I'm a bit confused. Yes, I know how to apply Carnot's theorem or the sine one, it's just that I don't understand what A and B refer to, I've read it to no avail. |
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07-15-16, 10:08 AM | #3 | |
Grey Wolf
Join Date: Jan 2014
Location: 50.1° N, 14.4° E
Posts: 834
Downloads: 82
Uploads: 5
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Quote:
Relatively distance target float is the same in relevance to target. The other quetstions are pure simple trigonometry, just draw the bearings, target course and even your course to new position to map, 4th. bearing from there and try to understand what side or angle of triangle you need to solve. The law of sine is ultimate solution for every sides and angles in that triangles.. |
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07-16-16, 07:35 AM | #4 | |
Mate
Join Date: Jan 2016
Posts: 58
Downloads: 13
Uploads: 0
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