Anybody in for a deep math question regarding fire angle?

Marcus · 10-11-05, 02:03 PM

Hi all,

Call me stupid but I want to solve the fire angle mathematically. Just for fun.
I do this with known variable True Bearing (from North), speed target, speed torpedo and course target. The true fire angle is from north and can be computed into the gyro angle.

Now I know I have the target pinpointed at X and Y starting from x= Range.cos (true bearing) and y= Range.sin (true bearing) going to X = Speed target. time.cos (course target) and y = speed target.time.sin(course target). I have the Aob. The (true) fire angle can be used to mark the place of interception x = speed torp.time.cos (true fire angle) and y = speed torp.time.sin(true fire angle). At interception both X's and Y's must be equal. This givve two equations with two unknowns, thus solvable. But I cant get to the solution though. Anybody want to take a shot here? It is driving me mad.
thnx in advance....

10-11-05, 02:03 PM	#1
Marcus Sailor man Join Date: Sep 2005 Location: Netherlands Posts: 46 Downloads: 0 Uploads: 0	Anybody in for a deep math question regarding fire angle? Hi all, Call me stupid but I want to solve the fire angle mathematically. Just for fun. I do this with known variable True Bearing (from North), speed target, speed torpedo and course target. The true fire angle is from north and can be computed into the gyro angle. Now I know I have the target pinpointed at X and Y starting from x= Range.cos (true bearing) and y= Range.sin (true bearing) going to X = Speed target. time.cos (course target) and y = speed target.time.sin(course target). I have the Aob. The (true) fire angle can be used to mark the place of interception x = speed torp.time.cos (true fire angle) and y = speed torp.time.sin(true fire angle). At interception both X's and Y's must be equal. This givve two equations with two unknowns, thus solvable. But I cant get to the solution though. Anybody want to take a shot here? It is driving me mad. thnx in advance....