120 Degree Track?

[groomsie]

With a new job I've had more time for leisure reading. I'm currently reading "Wahoo" by Dick O'Kane (which by the way does not seem as well written as I recall "Clear The Bridge" being--he seems to skip some details which I'm sure he knows the backstory but the reader may not).

Anyway, one item that jumped out at me was several references by O'Kane to Morton preferring a track of 120 as it afforded the target the least chance of evading the torpedo. I've used the "Dick O'Kane" targeting method with great success; I've played around with the "Cromwell" method when I have time and inclination and can appreciate that it provides least time to avoid (ie: faster rate of closure between weapon and target).

Has anyone tried anything like what I'm referring to (weapon approach aft of the beam around 30 degrees)? It would obviously lessen the closure rate between weapon and target but I hadn't thought about (and have not worked through) the thought that the angle may make evasion more difficult. Morton obviously was a really aggressive and innovative skipper who enjoyed great success, so I have to feel he knew what he was talking about (although the differences in game play--like dragster-like targets--may negate what in real life may in fact have been valid).

Anyway, I don't recall ever seeing this approach ever mentioned and was curious if anyone has looked into this.

[Pisces]

Are you sure that 120 isn't supposed to mean the torpedo track makes a 60 degree angle with the bow, and 120 with the stern. Because that way the torpedo still comes somewhat head-on-ish and the closure rate is bigger than the other approach.

What might have been the reason for it's succes in real life is that the torpedo's bubbles might have been disguised by the target's wake/bow-waves.

[groomsie]

As I completed my reading I also see reference by O'Kane to the 90-degree track, so it seems both were used. Here is where I got the 120-degree reference (I saw it several times, but this was the first I found to lend context); from "Wahoo: The Patrols of America's Most Famous World War II Submarine" by RADM Richard H. O'Kane, East China and Yellow Seas Chapter 4 (page 202 in my edition):
At 0455 it was light enough to see through the scope, and finally two blasts took us down, pausing while a final radar range and periscope bearing fixed her position. Both Chan and Richie agreed on an enemy speed of 9 knots instead of 10. Captain Morton had readied two tubes forward and aft, and now ordered the forward doors opened; it would be a bow shot. Her port angle was opening as it should. I could hear Richie advise, "750 yards from the track," as I called 90 port.

The captain waited for an approximate 120 track, the aspect that would allow maximum enemy maneuvers and still insure that our torpedoes would hit, and then said, "Any time Dick. Fire just one torpedo."
So, I feel confident that this is a torpedo track of 120 degrees off the port side (or a 240 target angle). Clearly they waited after the 90-degree bearing to allow the target to move beyond that point to fire.

Now, I did see where Morton also used the 90-degree track and whatever else came up for opportunity, but the comment that this would allow maximum enemy maneuvers with best chance to get a hit caught my eye.

Discussion?

[groomsie]

Same book and chapter as cited before, this time page 210;
Our approach to the firing point moved swiftly, in part because the enemy ship was cruising at 11 knots, and because of our compensating speed. Again, the captain selected a 120 track. That meant Wahoo's torpedoes would strike from 30 degrees abaft her beam, and I made a mental note for the future to accept any broad track and hit before another zig...
So, that entry removes doubt of the torpedo track. My questions remain as:


Confirming in my mind that this would be the most difficult track to maneuver away from (could play into advance and transfer of a ship maneuvering in real life, which I'm not sure how well the game simulates), and
Would this track work as well in the simulation (I'm guessing not unless one of the acceleration corrections has been added via a mod).

Thoughts? If nothing else I'd think Rockin' Robbins would have some insight...

[Pisces]

Oh, ok then. If the master says he does it like that... But I'm not sure I agree with his reasons. :arrgh!:

[Nisgeis]

Here is the fire control manual from the 1950s for fleet boats:

http://www.hnsa.org/doc/attack/index.htm

Thge optimum track angles are shown in plate XVII and plate XVIII. What calculations that graph is based on isn't shown. It appears from the graph though, that for a zero gyro angle shot, the maximum deflection angle you can get is the best.

That graph is a theorhetical example as it's for a zero gyro angle shot with a torpedo run length of 1,000 yards, so it's not something that you could manouver for. So, trying to read that graph as it is is quite hard.

If though you are shooting from behind, the speed of the torpedo will be reduced in relation to the target, which will mean that the target and torpedo have more chance to converge, as the torpedo will spend longer travelling across the target's course so the effective target length is larger. The opposite of the Cromwell method, where the target speed adds to the closing speed.

BUT! That graph is misleding, as you cannot directly compare a shot with a track of 45 degrees and a shot with 120 degrees, as you will need to be in completely different positions to get the same theorhetical 1000 yard run and zero gyro angle. Obviously you will need to be much closer if shooting from behind than if you are shooting from ahead to achieve a 1000 yard run length for both shots and the range difference in the two firing positions will increase the faster the target is.

[Hitman]

EDIT: Removed as I was wrong!

[groomsie]

Please re-read the quote I lifted from the book, posted at 12:22 yesterday.

It is clear that the Wahoo waited as the target passed 90-port to obtain the 120-degree track that Morton felt gave the enemy the least chance to evade. If it were a 60-degree angle as you surmised, they would shoot prior to getting the 90-port angle (as I typically do in SH4).

[EDIT: Removed some verbage to help clarify the thread.] I'll settle for a few links:

http://en.wikipedia.org/wiki/File:FiringGeometry.png (this from the TDC entry shows track angle, although it is listed as "theta subscript track").
http://www.history.navy.mil/library/online/ss-doc-4.htm (see 4107.i).

I think Nisgeis has hit it close to the mark, as that makes some sense. I may need to play around with it on paper and in the game to satisfy myself of how Morton/Wahoo were seeing the problem (and given their success I can't argue, although in the game it very well might be different...).

[groomsie]

I followed the link you gave and found the following on page 8-8 of the manual (pretty much (d) confirmed what Morton apparently did...I'll still need to work out the why...). Check this:

(c) It is within this range of torpedo track angles that the greatest amount of course error can be absorbed. From a study of the curves it is evident that the maximum deflection angle is obtained when firing on the optimum torpedo track angle and that the optimum torpedo track angle has a value equal to 90 degrees plus the maximum deflection angle. It is also evident that as the target speed increases for any given torpedo speed the slope of the curves becomes sharper. This means that the higher the target speed the greater the rate of change of deflection angle with torpedo track angle. It is therefore true that the optimum torpedo track angle is more effective for absorbing errors in course when the ratio of torpedo speed to target speed is large. It therefore may be stated that the optimum torpedo track angle is a good mean torpedo track angle for firing a salvo of torpedoes if the target speed is less than one-half of the torpedo speed. (d) The optimum torpedo track angle for a 16 knot target for a 46 knot torpedo is about 110 degrees and for a 29 knot torpedo about 125 degrees.

So based on this a 120 track seems reasonable. I would have thought something ahead (Cromwell style) would be best, but I suppose it may not be...

[Hitman]

I'm also trying not to be offended by the assertion I'm not familiar with what track angle means in the US Navy; I may be rusty but I've practiced fire control solutions and fired exercise torpedoes and missiles while serving in the same US Navy. I'll settle for a few links:

Ohhh my bad sorry. :oops:

I didn't know your background and thought that it all had an easier answer, please excuse my sillyness. My only justification is that I was trying to help and thought it was easier than it seemed. I'll delete my previous post to prevent confusions for other readers.

Sorry

[Pisces]

Ohhh my bad sorry. :oops:

... I'll delete my previous post to prevent confusions for other readers.

SorryTrouble is... it still is confusing if you can read the reaction to a 'supposedly bad' post, but not the post itself to make up your own mind. I guess keeping the peace is more important. :up:

[groomsie]

Didn't mean to come off quite so thin skinned, yesterday was a long day for me and I sometimes forget a wry smile doesn't come through in a thread. Also, I know it is easy to miss items in a thread.

Sorry if I came off a bit harsh, and thanks for adding your thoughts to the thread.

[Nisgeis]

It would help if they said what why it was optimum :DL. I can't quite work out what those figures are - e.g. a torpedo track angle of 0 or 180 with a deflection angle of zero is going to hit the target ship. yet on the graph it's the least effective. So, it must be a graph of true deflection angle (as you can't really mess that one up) versus estimated or judged torpedo track angle derived from your observation of target course. At least that's the only way I can make sense of those figures.

I can see how an up the kilt or down the throat shot would be the worst possible option, if there was course error introduced. This is where I can't work the graph out though, if they are taking into account course error, then are they taking into account speed error? The description would suggest that they aren't.