(Yet another) Attempt to make sense of gyro angles

[Von Due]

WARNING. MATHS AHEAD!

Introduction:

I am looking into the torpedo triangle problem where the gyro angle is not zero. As the game documentation offers no information on dimensions needed to work out the solutions, this is going to involve a lot of guesswork, trial and error. Some guesswork may be more educated than others but it will still be guesswork.

The most apparent problem is that SH3 is a game, that is nothing more than computer code and if the developers decided to use other dimensions than the historical accurate ones, or other principles entirely, then historical documents can only highlight issues without being able to offer actual values to put into the various formulas. This is where trial and error comes in, to attempt to deduce which dimensions and principles the game is using so that the non zero gyro angle triangle can be worked out by the player.

As a starting point, when it comes to errors, everything about the game is a suspect until proven innocent. To get anywhere one must make a few choices and assumptions. I will start with the historical accurate data (or what I believe is historically accurate) whereever I can find them, and attempt to work out if the results agree with what the game does.

To avoid this getting completely out of hand, I will limit this to the G7a and G7e torpedos.

Data is primarily nicked from http://tvre.org/en/gyro-angled-torpedoes

Finally, let me just say this entire topic is about torpedo shots where the gyro angle is not zero so no "just set it to zero, it's easier", please as that only makes the thread harder to read without adding anything useful.


General on torpedo runs as described in documents

When a torpedo leaves the tube muzzle, it will travel straight ahead for a short distance called reach (German: gerade vorlauf). This reach is torpedo type specific so when you know the type, you know the reach.
The reach is followed by the turn. The turn angle is the gyro angle. The turn also has a turn radius which is torpedo type specific so when you know the type, you know the radius.
The turn is followed by a final straight run to target. The gyro angle is the angle between the submarine's longitudal axis and this straight line. The distance between the tube muzzle and where the terminal run line crosses the submarine's longitudal axis, is called advance (German: Winkeleinsteuerungsversetzung). The advance is therefor a function of the reach, turn radius and gyro angle, two of which are constant for a given torpedo type.

The G7a and G7e

These torpedos have the same values for reach (R) and turn radius (r).
R = 9.5 meters
r = 95 meters.

The advance (A) for a 90* turn is reach + radius so for these torpedos, the
A = 9.5m + 95m = 104.5m

Generally, the advance for a given gyro angle (g) can be approximated by the formula

A = (R+r) (1 - cos g) according to http://tvre.org/en/gyro-angled-torpedoes

In the case of these two types, this gives

A = 104.5 (1 - cos g)

However, I will use
A = R + r * tan(0.5*g)

Periscope and parallax

The periscope is a fixed distance from the tube muzzle. This distance L is uboat type specific. The game offers no information on the precice distance here. Nor have I been able to find documentation on this value. This means that L + A can not be known. This value is necessary to solve the parallax problem. This may be decucable from in game tests though if a test method can be deviced that will give results accurately enough to being used. More than one method would be preferable to check for agreement between results.

Each submarine in the game has 2 periscopes some distance apart so L would depend on which periscope is used. The parallax effect has been confirmed on the two periscopes but the distance between the two is still unknown. To limit the work, I will only consider the attack periscope of the type VIIB uboat.

Restrictions on aiming

In the game, the gyro angle must be smaller than about 95* to either side of the bow (stern for stern tubes). For contact pistols, the angle between the torpedo path and target hull should be from 85* to 95* (from memory. If anyone has a correction, please post it) to minimize the chance of no detonation. For magnetic pistols, there is no such requirement.

Principles of aiming

When one aim the shot, what one is aiming at is a point P that is fixed relative to the submarine, at a constant bearing from the periscope, at a constant distance. As the submarine maneuvers in the world, this point will follow the submarine so the bearing and distance remain constant.

The moment the torpedo launches, that P leaves a virtual mark Q on the map. This virtual point Q will NOT move with the submarine but keep its position on the world map. Q is a time freeze image of where P was at the moment of launch. It is important to understand that the torpedo will be set to, at some point in the future, reach Q. If the submarine moves during the shot, then P and Q will not be in the same spot and it is only Q that can be considered.

There are 2 bearings to P that must be considered. One is from the periscope, the other from the advance point ahead of the muzzle, which will be L + A meters ahead of the periscope. Because the periscope is some distance away from the muzzle, and because of the advance, the bearing to P from the reach point ahead of the muzzle will be different than the bearing to P from the periscope. This is the root of the parallax problem.

The goal of the gyro angle is to have the torpedo reach Q at some point in time. Q is a mark left by P so P is the intersection of the two bearings that is, the bearing from the periscope (Bp) and the bearing from the far end of the advance (Ba). When the distance to P from the periscope increases at a constant Bp, then Ba will decrease and vice versa for decreasing range. There is a non-linear relationship between range and Ba. There is also a non-linear relationship between Bp and Ba.

Since gyro angle g = Ba, if one can find a function of Bp and range D that reveals Ba, then one has also found g.

Setting up the game's TDC for testing game vs theory

If the target is moving, then it is not enough to know P and therefor also Q. So far I have considered P as the intersection of Bp and Ba but for a moving target, Q must also be where the target and torpedo arrive at the same time. Q must therefor be where Bp, Ba and target course intersect. Furthermore, for contact pistols, it is desireable to have Ba perpendicular to target course, +- 5* but for the sake of more completeness, I will extend the work here to angles outside that window.

If the uboat's course is parallell to target course, then one can set up P beforehand and maneuver the boat so that P tracks the target course. If the uboat's course is not parallell, then P needs continuous update for range and or bearing. For that reason, I will look at cases where the uboat and target travel on parallell courses. Later one can look into the problems with non-parallell courses.

Periscope bearing, TDC and gyro angle

At this point I will remind you that P is where Bp and Ba intersect and that since Ba = gyro angle g, P is at the intersection of Bp and g. Bp defines a line of sight from the periscope, g defines the point P on that line. This means that once you set the range and Bp for P, and the target speed set to zero in the TDC, the gyro angle calculated by the TDC is precicely the angle that defines P. If g changes, then so does P. The opposite however is not necessarily true. This means that for a given P, there exists only one g, but for one given g there exist infinitely many possible P. Which one of these is determined by Bp.

As long as the target speed is zero, then the lead angle will be zero. Lead angle gives a lead bearing Bl. Bl is Bp offset for target speed and relative course. The TDC will always assume the periscope points at Bl. By setting the target speed to zero, Bl and Bp are one and the same and Bp can be read directly off the scale.

By setting target speed to greater than zero, the TDC will change g under the assumption that the periscope is at Bl but by doing so, it changes P. To correct that, you must turn the periscope so that g shows the correct angle for P. This is the angle found at target speed equal to zero.

A quick recap: The TDC is set for desired range to P and by turning the periscope to the desired bearing to P (Bp) with target speed set to zero, the correct gyro angle is found. Next target speed is given to the TDC which will change g and thus P. The periscope is turned so that g is back at the correct angle putting P back to where we want it.

(more to come as I keep working on it, including more on lead angle and ways to figure out true torpedo runtime).

[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.

[bstanko6]

Maybe look in USN manuals. I forgot where it is but there
Is a website with historical records for the US navy with manuals.

US subs in WW2 probably had the same issues.

[Von Due]

Thanks for the heads up on those! Will look around and hoping it would be for solutions before their uber TDC entered service.

I am slightly worried now though. If no analytical method is known, then that could mean that it is either proven to be impossible, or it is an NP problem. In both cases, the search would have to end. That would not be what I was hoping for. "Luckily" I am a complete fool and will keep looking until someone smacks me over my noggin with the evidence that there really is no analytical method known.

EDIT: Returning to the same site I found the other data, I discovered that the distance L I was looking for was roughly 28 meters for the Type VII boats. My guess was 33.5. Reality and virtual game reality are 2 different worlds but I will try to put 28 meters to the test and see how the game responds to that.

Another thing I found there was a description on how the TDS's parallax solver worked (sub chapter "Description of the component for calculating the parallax correction" at http://tvre.org/en/torpedo-calculator-t-vh-re-s3 ). Perhaps there is something there that sheds some light on this.

[Von Due]

To put L = 28m to the test, and assuming the reach and turn radius in game were historically accurate or close, I did this test:

Using L = 28 meters, R = 9.5 meters and r = 95 meters.

L + R = 37.5m.

I drew a line on the map straight up for 3000 meter using Pato's bearing and range tool. This line represents the Bp line of sight to P at a distance of 3000m (the line OP in the illustration above).

From the 0m end of the first line, I drew a horizontal line line and marked a point approximately 37 meters from the start, again using Pato's tool and counting the pixels (the vertical line in the illustration, and marking the end of the reach measured from the periscope at 0m).

From this mark, I drew a line approximately 95 meters long straight north. At the end of this line I then drew a circle that now had a 95m radius (the circle and horizontal radius line in the illustration).

From P at the far end of the 3000m line, I drew a tangent line for the gyro angle. Pato's bearing tool showed this line having a bearing of approximately 2.5 degrees (corresponds to the angle i) and the gyro angle g being approximately 92.5 degrees.

This is a very uplifting result as this agrees within the inaccuracies of the drawing and measurements to about 0.1-0.2 degree from what the TDC gives me for the same bearing and range.

It does appear more likely now that the game uses actual dimensions and that one could construct on the map the triangle for gyro angles at any arbitrary Bp and D.

EDIT: Repeated the test with L being my initial guess of 33.5. Changed the position of the reach and the center of the circle accordingly.

The difference for D = 3000 meters was too small to be meaningful but for D = 2000m the result was a closer match between the drawned angle and the TDC computed angle when L was 33.5. This suggests that 33.5 is closer to the length the game uses.

[Von Due]

Found this scan of a translated handbook for Kriegsmarine u-boot commanders
http://archive.hnsa.org/doc/uboat/index.htm#par148

It contains instructions on how to carry out various torpedo attacks including the 90* shot but also, curiously enough a 45* shot. Paragraph 148, II D IV.

It does not describe the calculations but it is an interesting read nontheless.

[La vache]

I am not able to reproduce all mathematical functions.
My source is: Analogrechner auf deutschen U-Booten des Zweiten Weltkrieges: Technikgeschichte und mathematische Grundlagen

The battle picture (Gefechtsbild) Tvh-Re/S3 similar Drawing 12/13 on tvre.org
The formula for the parallax improvement (Parallaxverbesserung) of Tvh-Re/S3

Sin (δ) = x/e * (ω + δ + Δω)