[wetwarev7]

Originally Posted by klh

When you're on a perpendicular attack course, you have formed a right triangle (from you to the target, from you to the intercept point on the target track, and from the target to the intercept point). The distances the target and the torpedo travel to the intercept point (where your torpedo goes boom) is related to their speeds. The ratio of the two (target speed / torpedo speed) is equal to the tangent of the lead angle (the bearing to the target at the time you fire). It is not necessary to know the range to the target, except to make sure your torpedo can travel that far).

It's just trigonometry!

See the Angriffsscheibe Handbuch, example # 16 for a more detailed explanation with pictures.

Wait a minute.....brain overload....rebooting....

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?

If so, why do I need to use a 20 degree bearing if I'm using a slower torpedo?

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....