There is a maximum bearing in the sense that larger and lower than 90/270 bearing produce less of a speed component across the line of sight. With the target on your beam (90 degrees to your bow) the speed component across the line of sight is the largest. So if your speed is not sufficient to match the 'speed across' of the target while it is on that bearing then you'll never get a constant bearing. It's going to outflank you and the bearing to the target will slowly drift to your bow.

The target speed across the line of sight are equal for AOBs that are equally far from 90. So a target closing with AOB 60 will provide the same speed across the line of sight as it would with 120 degrees moving away. AOB 30 would be equally fast as AOB 150 moving away. The difference with less than 90 degrees AOB, as opposed to larger than 90 AOB, is that the 'lesser' will have the bearing drift rate increase due to closing distance. Whereas the 'larger' will have this drift rate slowing down, due to increasing distance. So if you cannot match the speed of the target across the line of sight when it is on 90 then you should find other ways to speed up or forego this method. Then the Auswanderungsverfahren will work better, as it relies on the bearing drift to get a result.

As far as gyro angle goes, I do not immediately see your point with the later part of your example. A wide gyro angle does require a roughly accurate range setting. But only at the closest of ranges will this torpedo turn parallax will it have the largest effect. At close range the target appears the largest in size, so it wil somewhat mitigate theses chances for a hit. But I would not fire without some roughly accurate value for range to correct for the torpedo turn offset curve.

P.S. A bearing on 40-42 is not constant. I doubt you had a constant bearing solution for this method to work. Also a moving away AOB is a clear indication you are firing from a disadvantageous position. You better re-position yourself for attack.

If you are shooting with a gyro angle appreciably away from 0, say outside 20 degrees or so, an input for range becomes necessary to correct for parallax (the effect of the distance the torpedo travels before turning and the overall turn radius, in other words for the fact you aren’t shooting out your periscope). To do this Ausdampfverfahren As a shooting method, therefore you need to turn until the Gyro angle is close to zero.

The more practical application of this method though is simply using it to compute target speed. So instead of setting AOB to 90 and TDC speed to your speed, you simply establish the constant bearing, then multiply own speed by the sine of that constant bearing divided by the sine of the angle on bow and you’ve got target speed. Albeit that requires an AOB estimate.

The disadvantage of using this purely as a shooting method to derive a gyro angle to turn to is the fact that it’s only valid at that AOB. As soon as the target’s angle on bow starts growing the solution becomes invalid, so you need to turn quickly to that zero gyro and shoot in order for this to work.

Pisces I now see you’d already posted. Great explanation!:Kaleun_Cheers:

There is no reason to forego a gyro angle greater than zero degrees.
If you do not calculate the distance to the target for parallax correction with TDC, the parallax is compensated with the hit point at the target. There were tables with indication of shooting angle and target direction for parallax correction.
You can find it in my video:
Torpedo shot without fire control system.


https://www.youtube.com/watch?v=H2lxmdZvqHI

You are certainly correct. U-boat commanders handbook gives a value for an aimoff correction of 50m for a 45 degree angled shot. Woukd be greater for a 90 degree angled shot of course. Thanks for pointing that out!

Shoot... I am currently on a bad weather stretch right now, so most of my interception are through hydrophones until I reach a visual of < 1000 meters.

Well, you can do a makeshift intercept with the ausdampfverfahren just to get in range. Since you cannot see the target you cannot measure the AOB. So you cannot figure out it's actual speed and course. But if you try to manoeuvre your boat to get the target on a constant true bearing (heading) you have made your intercept. So at each hydrophone check look up the bearing on the white ring of the station, or add relative bearing to current course. Chase the target bearing drift until it is constant. If the bearing drifts away from course, turn towards it a bit. If it draws to your course, turn away a bit.

You wouldn't want to do this submerged for the hole run. Your batteries would be dead before you make visual contact. (aside from that it would take ages and may never even happen) Surfaced with own speed at 18 kts you only need to lead your course by 30 degrees max for the faster targets/convoys (9kts). Proportionally less for slower targets if you expect that to be the case. Doing hydrophone checks will slow down your average speed of course. So you may need to adjust for that. lead_angle= inverse_sine( expected_target_speed/average_own_speed) But with the distances in Wolfpack (edit: whoops, I forgot this is in the SH3 section) involved you may not need to do many hydrophone checks.

Still, shooting torpedoes based on that is out of the question. The range is too far and you can't see what you are aiming at. Besides, you don't know yet if it is moving away or towards you.

In the video to the "Ausdampfverfahren" below there are no calculation with sin. The own speed represents the target speed. Here its juts the own bearing from bow = AOB.

So with the formula it would be: Vt = 4 kt x sin(52°) / sin (52°)
So simplyfied its own speed = target speed when you are on a collision course

Why does this simplyfication work? And why should you use the formula instead?






https://www.youtube.com/watch?v=jLTkxym1y9s

Two ways to go about it:
1. You can set the angle on bow to 90 right or left, the consequence of this being of course that you need to multiply your own speed by the sine of the bearing, setting that as the speed.

2. Do as in the video here, which is probably easier to be honest, and that is to set your own speed as target speed, and set the angle on bow to the bearing.

In number 2, you are letting the TDC do the trig for you.

Target speed = own speed x sine (bearing) / sine (AOB)

So you can see by that formula, since the sine of 90 = 1, why you would need to do the math for number 1. In number 2 what you are doing is setting AOB equal to the bearing, therefore eliminating that side of the equation ( sin(brg) / sin(AOB) then = 1) so you are left simply with own speed = target speed.

But let’s take a step back. This method is good and it can be used in a pinch, but it’s not perfect since you need to be at a relatively small angle on bow in order to achieve the constant bearing if approaching underwater.

A much better use for the above formula is actually on the surface after you first sight the target. Turn to what you think is parallel, and adjust own course and speed at long range until the bearing doesn’t change over a few minutes, estimate the angle on bow, then apply the formula. Then once you get a better appreciation for the course, refigure what the AOB would’ve been, go back and drop that AOB into the original formula and voila - target speed by Ausdampfen, just like in the KTBs!

Edit: Forgive me because I see you realized the same thing about that side of the equation falling off. Use method 2, it’s easier if you’re going to use this method for shooting.

Thanks a lot!



Grüße aus Deutschland (Ich habe gesehen, dass du auch ganz gut deutsch kannst ;-)

Danke sehr! Fette Beute!:Kaleun_Cheers:

Please could anybody explain how to do the math with attack disc? I can multiply the speed by sine of the angle, but not sure how to do the trick with 3.2967 thing in the second method.

Error in Formula d
 

I think there is an error in the formula which computes d.

It is stated as d = E (in hm) x 3.29 (6000/1820) x sin(w)

But E can be given in meters. The 6000 is really 60, one hour x this change of angle and 1820 should be 1852, which is one nautical mile, to get from meters to knots.

Therefore the formula should be (IMHO)

d = E (in meters) x 60/1852 x sin(w). The differences are very tiny though...