 |
| From DE Docent 5K Walk |
I gave my 1st tour of the season. The tour was on the East shore of the river. The construction crew was putting the finishing touches on the new mooring, so the ship was delayed in moving from their winter berth (on the Rensselaer side of the Hudson River).
The group was a composed of Engineering Students, so I had an opportunity to discuss the mathematics of Naval Fire Control. It appeared that they had trouble with transportation earlier. When the bus appeared they were anxious to get aboard before the first group. I rushed the final three stops and got them on the bus. Jack Madden's group lost the race (I don't think they realized there was one), but Eric gave them they got a tour of the Motor Room and Engine Room while they waited for the shuttle bus to return.
Regarding math and Naval Fire Control here is a good example of the mathematics involved: ".... A differential equation is necessary to solve the problem of predicting the location of two maneuvering ships because of the changes of course they can make; only in an ideal situation will both of them be moving in a straight line.
Consider the following example. A firing ship is steaming course 000 (North) at 12 knots. Its target is steaming course 315 (Northwest) at 12 knots. Initially, the relative bearing to the target ship is 090 (East) and the range 10,000 yards, just shy of 5 nautical miles (4.937 nautical miles, to be more exact).
If neither ship turns, then their relative positions five minutes later may be calculated with trigonometry. The firing ship will have moved 1 nautical mile along course 000 and the target will have moved 1 nautical mile along course 315. We know that it has therefore moved 0.707 (sin(45) = 0.707) nautical miles along course 270 (West) and 0.707 (cos(45) = 0.707) nautical miles along course 000. Therefore, the distance between the firing ship’s track and the target ship has decreased to 4.230 nautical miles and it has fallen behind by 0.293 nautical miles. These distances form the two sides of a right triangle, with the hypotenuse being the linear distance between the two ships.
Trigonometry and the Pythagorean theorem again allow us to determine the length of this hypotenuse and the angle of bearing of the target ship This distance is 4.240 nautical miles ((0.293)2 + (4.230)2 = (4.240)2), and the angle is 86 degrees (arctan (4.230/0.293) = 86). As this is the angle between the firing ship’s track and the target, which is now slightly astern, the relative bearing becomes 094. The distance converts to 8,588 yards.
This simplest of examples illustrates the basic principles involved in plotting the movement of two ships. A significantly greater magnitude of complexity would be introduced if the target ship were turning to come onto a parallel course to the firing ship An accurate calculation would then require a differential equation. Effectively an infinite series of trigonometric calculations like the example above would have to be calculated as the target ship traced through the arc of its turn. ...."
A few more relevant links:
No comments:
Post a Comment