Anybody in for a deep math question regarding fire angle?

[Marcus]

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....

[Sturgeon]

Your approach will get to the right answer, but it is the hard way to do it. (There is one problem with your equations. The first equation for the intercept point should be x=x_start+Speed target.time.cos(course target) and similarly for y.)

The easy way is to draw a triangle. Draw a line from you to the ship's current position. This side has length R (the range to the target). From the ship, draw the second side of the triangle in the direction it is heading. It will be hit by the torpedo after 'time' seconds, so the length of this side is 'Speed target.time' (kind of like what you wrote). The angle from this side of the triangle to the first side of the triangle we drew is the angle on bow. The third side of the triangle is the line from the uboat to the intercept point, which has length 'Speed torp.time'. The angle between side 1 and side 3 is the angle offset (I call it theta) that the torp should take (measured from the current direction to the ship).

Now apply the Law of Sines, which states that sin(angle on bow)/(Speed torp.time)=sin(theta)/(Speed target.time). Solving for this gives sin(theta)=sin(angle on bow)*Speed target/Speed torp.

Notice that this solution does not depend upon the current range to the ship. (If you have heard of the term before, this is the 'proportional navigation' solution.) In SH3, the firing solution does depend a little on range, since it takes into account the fact that it takes some time and space for the torp to turn to the right heading. However, the solution doesn't change much with range for long range shots.

I hope this answers your question.

Or was this a geek test?

Sturgeon

[Marcus]

Thnx Sturgeon. I had the offset of the target calculated as you stated and like (LOL) you I called the "fire angle" theta. Already noticed that Range dropped from the equation but was unable to solve the remaining one. To be honest, forgot all about the law of sines! Thnx a million for your help. I drew up a spreadsheet on my PDA to manage the whole thing manually. Actually more fun then lock and fire. :D

[don1reed]

Hello Marcus,

here are two diagrams that may be useful:

The first one shows the Offset Angle of Collision (OAC) 16°, (assuming 40nm torpedo speed and AOB of 90°)

SS=Ship's speed
TS=Torpedo speed

http://img416.imageshack.us/img416/6217/gyrofiring5yw.th.jpg (http://img416.imageshack.us/my.php?image=gyrofiring5yw.jpg)

This second one shows the Kriegsmarine's diagram accounting for forward throw of the torpedo and 95meter radius gyro turn of the torpedo; and the paralax of the difference between the scope and the bow tubes.

http://img416.imageshack.us/img416/5159/tdcvinklar4dt.th.gif (http://img416.imageshack.us/my.php?image=tdcvinklar4dt.gif)

All the best,