OK here is a rather lengthy summary of my findings and my progress thus far. First, I used the
Silent Hunter II Manual Appendix C as my main reference, plus what little I can remember of high school trigonometry. If you read that Appendix C you will soon realize all you really need to do is solve the equation to find
a (the proper gyro angle)...
Target Speed V(t)
Torpedo Speed V(s)
Observed Angle on the Bow (8-)
Using the above equations we can solve for the proper gyro angle (a) with:
a = sin-1 ( (sin
* (V(t) / V(s) )
** sin-1 is the inverse sin function **
Method 1:
Solving the Triangles:
My first attempt dealt with solving for that angle of
B using trigonometry after all, if you manage to find
B the equation will give you
a. I quickly drew out the triangle on paper that is formed from the lines that are the Bearing to Target, the Target's Course and the Sub's Course. Using that triangle, you can solve for 2 angles and find out the reminaing one to find the correct angle
B, remember all the angles in a triangle add up to 180 degrees. Using this method I determined several equations I thought were nice rules of thumb to use in each situation. Situations like the target was 0->90 bearing to me.. or 180->270 bearing to me.. etc. etc. I thought it worked, and it turned out to be about 12 equations (fewer equations is always better) but, it looked like it worked. It didn't though. After double and triple checking the equations, specifically looking for ways I could make them fail, I found them. It was quite easy actually, the equations failed horriblly as soon as you start rotating the target around the sub's position. It turns out there aren't just 12 equations there were actually more like 256 or
more equations doing it this way. Obviously this is impractical, so on the subway home from work I had a flash on inspiration which led me to ....
Method 2:
Its all about the target....
The second attempt began with the premise of keeping things ultra simple. No lists of fancy equations, no multiple instances of different equatuions for variables, just keep things nice and simple. The real trick to this is realizing what that angle
B is supposed to be. That angle, the angle of the bow
B, is really nothing more then the bearing of the sub to the target. Pretend you are the target instead of the sub, what bearing is the sub from the target? Thats all we need to find. Onto an example...
Here we can see that
Bt (bearing to target) is 30 degrees port off the course of the submarine. We have no idea what
B is supposed to be yet but, we can find that using the reverse bearing (the bearing of the sub in relation to the target) and the course of the target. So first: Find the reverse bearing or also known as the bearing of the sub to the target. We need to know the true north bearing to the target from the sub to start with, not the mere 30 degree relative beearing. In the diagram we see the
relative bearing is 30 degrees and the sub heading is 78 degrees. So the true north (actual) bearing is 48 degrees.
Now we find the reverse true north bearing (
actual):
If this was ever over 360 then we just subtract 360 because you can't have a bearing that is more then 360 degrees, 360 degrees is a full circle.
Now we find the
relative reverse bearing: which includes is found with the
actual reverse bearing and the target heading. So to find that we find the difference between the target heading and the
actual reverse bearing,
So we now have...
Thats it. We have that
B angle we needed to find to solve for the right gyro angle (
a).
To make sure I'm not completely on crack lets see what happens when we turn the course of the target around.
Code: Select all
1. Bt = 78 - 30 = 48
2. Actual Reverse Bt = 180 + 48 = 228
3. Relative Reverse Bt = 243 - 228 = ?15
So, in conclusion, you can see from here that what is needed is that relative bearing to the sub from the target. Using these simple formulas and some logic you can find it in no time. I am working on setting these into some sort of nice equations that a calculator or other program will accept easily. But for now you can see exactly what the 'tough' part of this problem is and how you need to get it.
Feel free to ask questions or such if any of this doesn't make sense.
Tx again
, I was working on it too, but using vectors....but your method 2 rocks! I'm trying to program it for my Ti, I'll tell you about it when its done!