Quote:
Originally Posted by Rockin Robbins
And as a bonus, I found something else. You can put your target abeam and eyeball a relatively parallel course, right? You don't have to be accurate: 25 degrees off one way or another still yields pretty good results. Some idea of range is necessary too. But for what it's worth:
OK, that means that you're on a parallel course with your target. Your courses are the same, but of course, your speeds are different. We'll just say he's a mile, 2000 yards, away. Taking a series of bearings, his bearing is decreasing, he's falling back of 2 degrees per minute and you're going 10 knots. So he's traveling two knots slower than you are! Peg that guy at 8 knots. |
I could not tell from your post whether these were your words or someone else's words you were quoting.
I do not believe that would work. It is really designed for a stationary observation viewpoint where it is simple trig function to calculate the angle. Hence the warning about the statement disregarding the movement of the submarine.
This is because, if stationary, you are constructing a simple triangle with the submarine at one of the vertexes and the ship movement forming the two end points of the opposite side of the triangle. Knowing the relationship between the angle changing over the period of observation and the range, it is possible to calculate the speed. However at ranges over 2,000 yards one would either need to be able to measure the bearing to 0.01 precision or extend the period of observation for longer (10-15 minutes minimum). This can be demonstrated using Excel with the formula =DEGREES(atan(b/a)) where b= the target ship and a= range.
All this works great at long as the submarine is stationary. Once the submarine moves the simple triangle problem becomes a very complex quadrilateral problem.
I just don't think we would ever have enough information to be able to solve a single quadrilateral problem like this. I am trying to work out whether a series of contiguous quadrilaterals might be workable but so far it does not look good.
If, and this is a large if, I can steer my submarine on a parallel course with the target ship, the problem becomes a trapezoidal problem. But the only immediate payoff of that would be that I could calculate the target's AoB and, therefor, the target's course.
AoB = 180 - bearing to target (expressed as a number 0-180 not 0-359). Note, this
only applies when the submarine is on a parallel course with the target.
If, I somehow know the target's speed, I can take two bearings over a measured interval and calculate the range...this is not a simple formula but a combination of separate geometry and trig calculations. I am trying to find them and baring that I will try to create it.
Surprisingly calculating the range using this method is pretty straightforward if you use plotting paper and scale drawings, but trying to find a series of equations is challenging.