Q. Have you reviewed this trim and stability booklet prior to the testimony today?
A. I would not characterize it as a review. I have seen it. I looked at it.
Q. Do you know if there’s any GZ curve, minimum required GM. So how is a master supposed to understand all this about GZ and righting arm, clearly from a master’s standpoint, how is he supposed to – he or she supposed to look at this and say,okay, I need to weigh the factors, GZ – looking at this trimand stability booklet, or any of the operating instructions in here we see a curve – we see a minimum required GM curve and – have you reviewed any of the CargoMax forms for the vessel.
A. I have not.
Q. Would you also look at that table? And there’s a white box embedded in the table where it says minimum required GM values in this diagram must be maintained for all operating conditions to meet – and this is the important wording to me – weather criteria as specified by the U.S. Coast Guard. Do you see that?
A. Yes, I do.
Q. So this manual here, what weather criteria – is it a wind – is wind involved in this? Is sea state involved in this weather criteria that this manual was approved by ABS under – I believe recommendations should be included in there for the Coast Guard or we’ll deal with that in a second. Can you elaborate on this weather criteria for us?
A. Yes, it’s – I believe the weather criteria in 46 C.F.R. Subchapter S, 170.170, where we have a notional steady wind, I believe in the order of – or it translates to about 55 knots for open waters. The formula doesn’t say 55 knots it’s based on projected lateral area and length of the vessel, but essentially it is that. It is steady wind speed, and a minimum GM requirement based on that criteria.
Q. So it’s not talking about GZ here. It’s talking about GM – minimum required GM for certain weather criteria. Is that correct?
A. Yes.
The El Faro was lengthened in 1992-1993 which was ruled a major conversion. SOLAS 90 would have applied as it was effective in 1992.
The then renamed International Maritime Organization (IMO) adopted these probabilistic subdivision and damage stability regulations for dry cargo ships
which were then incorporated as part of SOLAS 90 Chapter II-1, Part B-1, which became effective for all cargo ships over 100m in length in 1992. SOLAS ‘90, following the Herald of Free Enterprise casualty in 1987,
The conversion to containers (RO-CON) was 2005-2006 but the 1990 stability standards would have applied.
From the CG report p -183
1990 SOLAS damage stability standards would still have been applicable in 2006, since
the 2009 SOLAS damage stability standards requiring a higher level of safety were not
applicable until 2011.
Presumably46 CFR 170.170 is the equivalent to SOLAS 90
From there, 170.170 is ; 
PAH being Pressure x Area X height - W is displacement and T is the angle of heel (1/2 Deck edge immersion or 14 degrees. - The pressure being approx the equivalent of about 55 kts.
Looking at this again:
At 14 degree list the the GM (slope of GZ curve) is roughly the same initial GM but falls off quickly after about 18 degrees or so.
Read for P = .055 (L/1309) 2 metric tons/m 2 in § 170.170: P = 0.055 + (L/1309)² metric tons/m² otherwise you end nowhere with a calculation!
As an example with a couple of rough estimates:
- P = 0.088896
- A = 2880 m²
- H = 10.5 m
- W = 38490 m³ (Fully loaded, draft = 9.2 m)
then the GM = 0.28 m.
Yes,from .gov its " +"
P = .055 + (L/1309)^2
| L | 240 | m |
|---|---|---|
| A | 2880 | m |
| H | 10 | m |
| W | 38490 | ton |
| T | 14 | degrees |
| Tan T | 0.249 | |
| P= .055+((L/1309)^2) | 0.089 | |
| PAH | 2552.134 | |
| W tan T | 9596.635 | |
| GM | 0.266 | meters |
What does this mean?
H = the vertical distance in feet (meters) from the center of A to the center of the underwater lateral area or approximately to the one-half draft point.
Draft was 30 ft or 9 meters 9/2= 4.2 H should be about 1/2 distance to top of container stack plus 4.2 meters?
H = the vertical distance in feet (meters) from the center of A to the center of the underwater lateral area or approximately to the one-half draft point.
Draft was 30 ft or 9 meters 9/2= 4.2 H should be about 1/2 distance to top of container stack plus 4.2 meters?
Yes, I calculated an average of 3,5 stacked containers. Container height is 2.60 m that is a total height of 9.1 m which figure I rounded to 11 m because of the higher house. Then H = 11/2 m + 9/2 m = 5.5 m + 4.5 m = 10 m.
5.5 meters is from load line to top of the container stack? plus a bit?
Yes. It is a rough estimate only and a bit arbitrary but it serves the purpose of getting a general idea about things. Interesting GM formula.
The fomular is senitive to angel T. It should be 14 degrees or 1/2 DEI so 14 is likely to high? What answer do we expect, a little over 1 meter I think.
| L | 240 | m |
|---|---|---|
| A | 2880 | m |
| H | 12 | m |
| W | 38490 | ton |
| T | 7.5 | degrees |
| Tan T | 0.132 | |
| P= .055+((L/1309)^2) | 0.089 | |
| PAH | 3062.561 | |
| W tan T | 5067.305 | |
| GM | 0.604 | meters |
| GM | 1.98 | Feet |
I concur, just over 1 m I think and that comes more or less into the direction of the 1.3 m of the final voyage. Well, alltogether it is just a interesting exercise.
For H the distance from the freeboard deck to the bottom of the containers is missing. That wold be the height of the 2nd deck which is RO/RO.
H = the vertical distance in feet (meters) from the center of A to the center of the underwater lateral area or approximately to the one-half draft point.
I see what you mean, this is a complicated ship!
I am getting some shuteye now, probably dreaming about it…
I get 14.3 meters from freeboard deck to the top of the containers. Freeboard is 12 feet or 3.6 meters 14.3+(3.6/2)= 16 meters
H=16 meters + 4.5 m
=20.5
Area would be 16mx240m=3840 m^2 ~ 3500 m^2
| L | 240 | m |
|---|---|---|
| A | 3500 | m |
| H | 20.5 | m |
| W | 38490 | ton |
| T | 7.5 | degrees |
| Tan T | 0.132 | |
| P= .055+((L/1309)^2) | 0.089 | |
| PAH | 6358.182 | |
| W tan T | 5067.305 | |
| GM | 1.255 | meters |
| GM | 4.12 | Feet |
I may have made an error somewhere but that brackets the problem, must be in the ballpark.
Do you mean the brackets as in P= .055+((L/1309)^2)? They are ok but mathematically you can leave out one pair: P= .055+(L/1309)^2 unless you multiply it like in GM = ((.055+(L/1309)^2)* A* H/W*tan(T)
I think he was speaking in terms of naval gunnery, where achieving one shot a bit short and one a bit long gives favorable odds for the third shot.
Not “that bracket is a problem…” but rather “that does bracket the problem…” 
Soooo…the upshot is El Faro met the requirements of 170.170? Because 4.12 was required and it had 4.28?
Oops, I didnot see that one coming, here my poor understanding of (gunnery) English is beginning to show…
This was a statement earlier made by Stettler.
In terms of stability, the El Faro’s metacentric height — a measure of stability, expressed as GM — met the 1990 Safety of Life at Sea (SOLAS) standard at 3.1’, Stetler reported. However, if a comparable vessel were built in 2016 the current GM standard would be 5.8’, Stetler wrote.
Your English is excellent, sir. I would take you for a native speaker. The lack of an apostrophe was a clue, but I think that many, perhaps even most native speakers would be unaware of that particular idiom unless they had an interest in military history or practice. For myself, circumstances led me to be reading WWII naval history age six or so, as my father fought in the war and the house was filled with it.
Yes the requirements (3.1 feet) were met. In addition to the legal minimum the ship used an additional 0.5 foot margin.
BTW the mate doesn’t need to make any calculations or judgment calls, the weights are entered into CargoMax and the results can be seen, either OK or not. Same with lashings and many other things, the crew on deck just follows the cargo and lashing plan, no need to have in-depth understanding.
Two items of interest here, one is how changes in vessel design can “beat” applicable stability regs. This is what happened on the EL Faro with the RO-CON conversion.
Here’s where OSV design outran the regs:
The other thing is that if the crew wanted a better GM margin they had to directly push back against the money guys and leave cargo on the dock. That’s because the ballast tanks were converted to permanent ballast so that the ship could be overloaded and still be legal.
In my experience on PCTCs we have much more flexibly, if we want more GM we just add more ballast. They run out of cargo space before we run out of ballast capability.
The stability of platform supply vessels is an interesting subject on its own. When recovering anchor buoys that on one semi submersible were 19 tonnes each the normal practice was to trim the AHTS by the stern.
An analysis of the stability data showed how dangerous this practice was and it was discontinued.
The Swire groups training centre in Singapore for officers in the offshore industry produced an excellent video which makes it crystal clear how the loss of stability occurs.