Naval History Forums

As some members might remember there is the riddle of the very cryptic draft measurements during Bismarck’s trial runs, fall 1940. We have discussed it at one time or another:

viewtopic.php?f=1&t=1828&p=21066
viewtopic.php?f=1&t=3142&p=27863#p27862

The meaning of the /4 was unknown, but usually the other value usually was considered as the draft in meters. In my opinion this was wrong, because there are values which make no sense. ...however, I know people who have considered them as typos.

Due to a fortunate coincidence I’ve found the meaning and the calculation rule yesterday:
The first value is the point of intersection of the waterline with the draft marks; the second value is the number of the visible bars of the draft marks.


Example: That means the draft measurement would be 2,3/4 in this case.

The calculation rule is:
position of the highest visible bar of the draft marks above keel+ the value of the point of intersection – number of bars x10=draft (all in dm)

In my example: 130 dm+2,3 dm-4x10 dm=92,3 dm ->9,23m

Marc,

Thank you very much for posting this.

So, if I understood that correctly, for the famous speed trial of 29 October, 1940, we have the following data:

Draft forward: 8.5 / 4 = 9.85 meters.
Draft aft: 8.8 / 4 = 9.80 m meters.

Mean draft: 9.825 meters, which corresponds to a displaced volume of approx. 47,900 m3 = 49,100 mt in seawater (1.025 relative density).

Bismarck speed trial, 29 October, 1940.

Displacement: 49,100 metric tons (75% Zuladung). Matches very well with a load of 6.285 m3 of fuel oil according to the KTB.
Water depth 65 m
Total power: 117,410 shp
Speed 28.374 knots.

Hope you having a great summer!

José

Wow, that is really tricky, why did they go to such complication? May be so the draught was not guessed at a foreign port? I always have the impression that German displacements above allowable limits were tolerated because it is relatively easy to guess a ships displacement knowing length, beam and draught, this system would hide one of the terms in the equation.

Regards

@José

8,8/4 = 98,8 dm = 9,88 m

@Marcelo

Well, it was called "getarnte Tiefgangsmeldung" (masked draft message). Without the height of the highest bar above keel and the calculation rule the values are quite worthless. IMHO it should prevent that any unauthorized reader (or even authorized bearer) of such a message would have easily access to speed and draft information. I think our cluelessness for many years is the best example that this was very effective.

Herr Nilsson wrote: ↑Tue Aug 20, 2019 7:03 am @José

8,8/4 = 98,8 dm = 9,88 m

Thanks, my mistake.
That gives a mean draft of 9.865 meters which corresponds to a displaced volume of approx. 48,000 m3 = 49,200 mt in seawater.

Let me try to understand this. There are 13 bars, I count how many are above water line, I take the next below water, in the photo would be number 9, so I start with 9 m, and add the decimeters to this. Right?

@Marcelo

Not exactly: there aren't all 13 bars, but a just couple of bars in the relevant area around the waterline, which indicated the distance over keel. Furthermore your sequence of calculation differs from the original one. You first read the decimeters of the waterline and add it to the decimeter-value of the highest bar and then you subtract the number of the counted bars over waterline in decimeters (->number x 10)...but the result should be the same.

That's a calculation of 3,8/2 and the highest bar at 11 m:

Yes, with a little algebra you can reorder the original formula to mine. The system seems to have worked well, because the contemporary Jane´s gave 28 feet draught for Bismarck and 24 for Scharnhorst.

My theory is that it would not have been difficult for the would-be Allies to assess the correct displacement of these ships. The dimensions listed for Bismarck were 791 x 118 x 28 feet, using a prismatic coefficient value of 0.55 those dimensions traduces to a displacement of 40700 t, almost 6000 t above allowable limit. For a ship with that dimensions to displace 35000 t, the prismatic coefficient needs to be 0.47, this would necessitate concaves waterlines.

Regards

José M. Rico wrote: ↑Tue Aug 20, 2019 12:49 am Marc,

Thank you very much for posting this.

So, if I understood that correctly, for the famous speed trial of 29 October, 1940, we have the following data:

Draft forward: 8.5 / 4 = 9.85 meters.
Draft aft: 8.8 / 4 = 9.80 m meters.

Mean draft: 9.825 meters, which corresponds to a displaced volume of approx. 47,900 m3 = 49,100 mt in seawater (1.025 relative density).

Bismarck speed trial, 29 October, 1940.

Displacement: 49,100 metric tons (75% Zuladung). Matches very well with a load of 6.285 m3 of fuel oil according to the KTB.
Water depth 65 m
Total power: 117,410 shp
Speed 28.374 knots.

Hope you having a great summer!

José

Gentlemen,
I'm afraid I am totally lost when it comes to calculations like this (Mathematics were never my favorite subject!) but are you suggesting that a recorded trial speed of 28.374 knots is actually less that Bismarck was supposed to be capable of and therefore very little if any more that of PoW?

Hello Paul,

although we knew the speed and the SHP's, we never had an idea at what draft/displacement. It was assumed that Bismarck made her speed in very light condition, because 8.65 meters (using 8.5 and 8.8 as meter values) is almost "empty ship". Knowing the calculation rule, now we know that Bismarck made her trials at a more plausible displacement. That means she already made ~28,4 kn at just 117,401 SHP and a displacement that approximates the 75% payload at which trials were usually made. But this was just one of a lot of trial runs and Bismarck also conducted another trial run at ~150,000 SHP at most probably 30.1 kn.

Fellow Contributors,

although we knew the speed and the SHP's, we never had an idea at what draft/displacement.

Gentlemen can we just confirm these "famous" speed trials took place in the Baltic?

All the best

wadinga

wadinga wrote: ↑Fri Aug 23, 2019 6:40 pm Gentlemen can we just confirm these "famous" speed trials took place in the Baltic?

Yes we can.
All speed trials took place in the Gulf of Danzig, in 65-meter deep waters off Neukrug, in the Vistula Spit, not far from Pillau (today Baltiysk).
Why?

I assume because of the density of the water.

Hello everybody,

Jose Rico wrote: "All speed trials took place in the Gulf of Danzig, in 65-meter deep waters off Neukrug"

Wouldn't these relatively shallow waters affect negatively the ship's speed performances ?
The Italian RN Roma could sustain her speed trials (August 1942) only once close to the Otranto Channel (where depth was well over 200 meters) as the Northern Adriatic Sea was not allowing a meaningful trial.
I couldn't find any relation/indication of how the depth of water affects a ship's speed, however...


Bye, Alberto

There are some writings of D.K.Brown about this, shallow waters (meaning about 10-20 m) allows more speed, I think that 65 m is out of the effect.

Regards