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#10 |
XO
![]() Join Date: Jun 2014
Location: Chorrillos, Lima, Peru
Posts: 401
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All right. Let's assume that we are firing a 28 knot torpedo at a ship that will be 2800 meters away from our original point of firing at the moment of impact. We also assume that the target will be at exactly 90º AOB at the moment of firing. Thus we have a right triangle with a hypotenuse of 2800.
Now let's say, for the sake of argument, that the lead angle is 20º – what will the target speed be? Sin(20º) = 0.342 so the ship will have traveled 957.7 meters from the time of firing until the time of impact. Accordingly, the ship is traveling at 9.577 knots. So your formula is: Lead angle = torpedo speed - ship speed ? 28 - 9.6 = 18.4 not 20. What if the ship is traveling more slowly? Let's try 2800 hypotenuse and 400 opposite (28 knots and 4 knots relatively). What's the angle? Well, if your formula is right, it should be 28 - 4 = 24? Not even close. The real angle is around 8.2º ------------------ Sorry. I don't get it. |
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