(Yet another) Attempt to make sense of gyro angles

[Von Due]

I am trying to establish the distance L between the attack periscope and the bow tube muzzles.

I set up P at Bp = 270* at 3000m

Totally eyeballing the terminal run back to the 0* bearing line to somewhere near 140 meters from the center of the submarine icon.

Assuming the game uses the correct values for reach and advance, I backtrack the distance L to be in the area of 33.5 m +- 2 meters uncertainty.

I will use 33.5 as my initial guess and look at the consequences, to try to narrow it down to a closer to exact value.

If 33.5 is correct, then L + R is constant and equal to 43m.

Then the side S of the triangle that is on the 0* bearing is

S = 43 + 95 * tan (0.5 * g)

[Von Due]

Made an illustration to the problem with gyro angles as a function of Bp and D.

[Von Due]

Digression: Assuming the developers used a similar model as in the illustration but ignored that port and starboard turns would require mirrored models, it might be possible that this would be the reason for the discrepancies between port and starboard turns in the game, yielding so different projections on the attack map. If this is the case, then the model would work for one turn but not for the other. One more on the things-to-check list.

[bstanko6]

I don't believe German torps could travel in a curve and then straighten out, being they were preset gyros.

Firing a torp is no different then a handgun. Trigonometry plays a huge
Role.

If I stand at the range and hold a gun, my hand naturally moves and shakes.

If I pull the trigger, the bullet will travel, and every shake and tremble will throw the trig solution off the greater the distance the bullet travels.

Germans knew this, and ordered their boats to fire under 1000 meters!

Also, setting the gyro on a torp to curve at a greater degree, is like pulling a guns trigger while moving... the degree of error is too great.

This is why we try to get a zero angle every time, to eliminate error.

You are doing a lot of work on this TDC. But I think due to German engineers lack of developing the torp and TDC compared to tanks and planes in WW2, you are better off eliminating error, than solving angle issues.

Torps cost money! Don't be wasteful!

[Von Due]

Originally Posted by bstanko6

I don't believe German torps could travel in a curve and then straighten out, being they were preset gyros.

Firing a torp is no different then a handgun. Trigonometry plays a huge
Role.

If I stand at the range and hold a gun, my hand naturally moves and shakes.

If I pull the trigger, the bullet will travel, and every shake and tremble will throw the trig solution off the greater the distance the bullet travels.

Germans knew this, and ordered their boats to fire under 1000 meters!

Also, setting the gyro on a torp to curve at a greater degree, is like pulling a guns trigger while moving... the degree of error is too great.

This is why we try to get a zero angle every time, to eliminate error.

You are doing a lot of work on this TDC. But I think due to German engineers lack of developing the torp and TDC compared to tanks and planes in WW2, you are better off eliminating error, than solving angle issues.

Torps cost money! Don't be wasteful!

Thanks for the input but as I stated in the introduction, this topic is solely for the discussion of shots with gyro angles other than zero and as such, any response here like "don't do that" is only making the thread less readable. If you read the page I provided a link to, then you would have known that according to that page, German torpedos, as well as most other torpedos after WWI could indeed turn. This includes the G7a and G7e. Not only were the torpedos capable, but the submariners including German ones, knew about it and knew how to solve it. However, for tactical reasons, only 90 degree gyro settings were used as alternative to zero degree runs.

[bstanko6]

Sorry I got lost in your MIT dissertation you wrote. Just kidding, I am being funny. I actually started to skim over and I did not read that part about certain responses.

Any way, my concern was over the statement you made in the first paragraph of your second topic:

"The turn is followed by a final straight run to target."

I did not think this is possible for an analog gyro system to do. I can see modern torpedoes straightening out, not WW2 era. That is what I really meant.

I added the trigonometry comments because it is true, you are sending a torp (bullet) with no controls once it is fired, onto a path that cannot be changed. The only thing that changes is your boat moving, the target moving, every wave and sea state your torp travels through, the density of that sea state changing, the resistance of drag onto the outer shell of the torp as it travels through said density, the distribution of steam (or power from battery cell) of torp, all the while you wanted to set a curved path with manual input that may or may not be correct depending on the math of the person who inputs the solution in the first place....

needed to breath there!

When in the end, you can eliminate the bulk of these variables by eliminating the gyro angle, and set it to 0!

[Von Due]

You must remember what the gyroscope is actually doing. Before launch, the gyro is set to an angle to zero degree bearing. The way gyros work, they will want to turn into this new course and stabilize themselves there once they are on the new course. The gyro setting is not a turn rate setting but a course setting, a course relative to the longitudal axis of the torpedo's pre-launch position.

Example: Your boat is on a 250* true course and you fire the torpedo with a 30* port turn gyro setting. That will take the torpedo to the course of 220* true heading. It will not continue the turn after reaching 220* because this is the heading where the gyroscope is not fighting back to correct the course. This is the course that makes the gyroscope happy.

If your true course is 315* true heading, then a 30* port gyro angle will see the torpedo take the true course of 285* true heading.

After this heading is achieved, no more turning takes place and with no turn, there is only a straight run left to do.

[bstanko6]

That's right. You are right! My apologies. I was thinking in an extreme curve where the torp hits the boat while curving.

Thank you for clarifying!

[Von Due]

Referring to the illustration in post #3.

The circle represents the turn. It has 2 tangent lines. The vertical tangent line is the uboat's 0* bearing line. The other going to P is the bearing line for Ba.

The turn starts at the tangent point at the 0* bearing line. The turn then follows the curve until it reaches the 2nd tangent point at the Ba line.

The Ba line must always be a tangent line to the circle. So must the vertical 0* bearing line.

This means the intersection angle i is a function of Bp and D.


The 0* bearing line, the Bp line and the Ba line form a triangle with angles
Bp, i and (180* - g). Regardless of the length of the sides in this triangle, any triangle with this particluar set of angles are all similar and can be treated as one and the same when solving for angles.

With Bp and D as function inputs, one can therefor describe the gyro angle g as a function of Bp and D.

Value of g = g(Bp, D)

The Quest now then, is to find that function...

EDIT: Apparently, German submariners would bring with them tables for parallax correction and I can only find references to numerical methods for estimating the corrections. I can not find any reference to any analytical method which is bad news.