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Originally Posted by wetwarev7
Wait a minute.....brain overload....rebooting....
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Be sure to reboot to SAFE mode.
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Originally Posted by wetwarev7
So, the farther out I am from the ships course, the farther out the target is going to be from the intersection at the same angle? So no matter what distance you are at, the firing angle(say a bearing of 10 degrees) will allways ensure that you and the target are at the correct distance?
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Exactly!
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Originally Posted by wetwarev7
If so, why do I need to use a 20 degree bearing if I'm using a slower torpedo?
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Think of it this way. A slower torpedo will take longer to reach the intercept point (the target's track), so the target will travel further during that time. Therefore, you need to lead the target more.
And don't get hung up on the 10 or 20 degrees given in the example. You can calculate (or use tables like the real kaleuns did) the exact lead angle with the equation...
Lead Angle = Arctan (target speed/torpedo speed)
So if the target is making 12 knots, and you wish to fire a 30 knot T-III electric, then your lead angle is...
Lead Angle = Arctan (12/30) = 21.8 degrees
And as long as you're within 5000 metres of the target track, you should get a hit!
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Originally Posted by wetwarev7
I don't know why this stuff confuses me. I aced trig in highschool, but maybe because it was so long ago, and I've never had a real world use for it till now....
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I keep telling my kids that this stuff will come in handy. Now I can prove it!
