I do not see how two ships can alter their bearings when stopped Lord Mersey


Paul Lee

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Another point to ponder:

2nd Officer Stone was adamant at the British Enquiry, that the "other" steamer he saw
started steaming away slowly to the SW at the time of the first rocket being fired. However,
it is generally assumed that he was mistaken, as Gibson (even though he had been off the
bridge for a while trying to ready a new logline for the morning) didn't see their strange ship
moving away.

However, looking at their accounts given to Captain Lord on the 18th April, Stone was pointed
out the other ship by Groves when he came on the bridge at 12.08am. The Californian was
heading ENE and the other ship was SSE, dead abeam. Gibson also saw this ship on the
starboard beam when he came on the bridge at 12.15am.

It is hard to reconcile Gibson and Stone's ship bearings from their accounts - for one thing,
Gibson was away for some little time, and he tended to give relative bearings (relative to
the Californian's head or beam), whereas Stone would give bearings by the compass (and,
as was said at the enquiry, he was constantly taking bearings of the other ship).

However, there is one point in their accounts that does allow their memories to be compared.
At 1.50, Stone recalled that the Californian was heading WSW, and the other steamer was
about SWxW; Gibson also made a note of the other ship's position, as he relayed Stone's
information to Lord in the chart room. Gibson saw the last rocket fired when the ship was one
point on the starboard bow.

Now, by Stone's account, with the Californian heading WSW, and the other ship bearing SWxW,
this would be S 56.5 W, or, as Gibson says, 1 point on the port bow. Now, it is possible that Gibson
was told the other ship was this bearing off the bow, but why would he have said "1 point on the port
bow", rather than SWxW in his written account.

How can a stopped ship go from SSE at 12.08, or whenever she stopped, to SWxW at 1.45am?

Cheers

Paul

 

Paul Lee

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I thought the current might have something to do with it, but surely it would have to be quite strong for such a change in bearing?

If the ships were 5 miles apart at 12.08 and 11.45, that would imply a movement of 6.4 miles. I suppose its possible that the Californian and Titanic were caught in a drift that made them drift further apart, and pushed them apart in an E-W direction. Might be possible with the convergence of the Gulf Stream and the Labrador Current...

anyone care to comment?

Cheers

Paul

 
Dec 2, 2000
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>>I thought the current might have something to do with it, but surely it would have to be quite strong for such a change in bearing?<<

Not necesserily. Remember that a ship out on the open ocean isn't really tied to anything so she's going to tend to move or turn in whatever direction the prevailing currents will take her. And when even a 1/2 knot current is backed up by the weight of the entire ocean....well...do the math!
wink.gif
 

Paul Lee

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Alright, but how can it make something move so much? If the 5 mile distance is right, then the other ship moved 6 miles in an hour (or 3 miles each, I'm not fussy). I thought the current that night was generally accepted to be between 1/2 and 1 knot to the south?

Cheers

Paul

 
Dec 2, 2000
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As I indicated, the current is backed up by the weight of the entire ocean. That's a lot of sheer mass working against a hull that is, for all practical intents and purposes, weightless. You really don't have to go way out to see in order to watch this sort of thing in action. Any harbour where there's an anchorage will do. The tides going in and out of a lot of them are not always that dramatic, but you can watch ships being swung around on their anchors in a reletively short period of time.

There's really nothing at all remarkable about this at all.
 

Paul Lee

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Hi Michael,
We're not talking about "swinging" though - we're talking about movement from a bearing of SSE to SWxW. I can see how a current may make a ship turn, but I can't see how it can make a ship change its bearings by so much in such a short space of time.

Cheers

Paul

 

Dave Gittins

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If a thing was physically impossible, it didn't happen. There is quite a simple explanation for Stone's observations. I'm not saying what it is but it involves a compass and a simple mistake that I've made myself, like a lot of others.
 

Paul Lee

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It wouldn't involve forgetting to take into account compass deviation would it? (The deviation that night was only 2 points).

Gibson does seem to back Stone up on this though. Although Gibson only took bearings relative to the Californian's head or beam, he does tell us what relative bearing the other ship was when she fired her last rocket - and knowing that the Californian was heading roughly west at the time, this can only mean that the other ship was SW (approximately).

Cheers

Paul

 

Paul Lee

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I've drawn a very crude chart to display the data, as they stand:

90078.jpg


Sorry if some of it is a bit illegible: the jpeg compression routine was perhaps a little bit over-zealous!

Cheers

Paul

 

John Knight

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I have been following this thread for some time and am impressed with the thought that has gone into it. Thanks for the chart Paul even I could understand it.
Regards,
John.
 

Noel F. Jones

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"As I indicated, the current is backed up by the weight of the entire ocean."

I cannot see that oceanic set will of itself swing a stationary vessel relative to a compass bearing. Her location will only be translated 'over the ground' without any change in her aspect. This is analogous to an object on a conveyor belt.

A vessel with the way off will answer to wave action, falling approximately broadside-on thereto. She will also answer to any disproportionate wind pressure on her deck erections, her final aspect being the result of some balance between those two external forces.

Feel free to educate me otherwise....

Noel
 

Paul Lee

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Thanks for the kind comment John. The only problem with respect to the sketch is that, for the geometry to be "right", the Californian's positions would have to be well to the east of where she stopped for the night - and I can't see how that could happen. Not even a strong drift could explain it, as the Californian was able to travel to the wrecksite in under two hours the next day.

Cheers

Paul

 

Dave Gittins

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Noel's explanation is a very good one. Captain Lord himself made it clear that his ship was turning due to the action of the wind.

Paul, why do you believe Lord's evidence on his overnight position? (By the way, Lord took 2½ hours to get to the boats and Carpathia. 6-00 to 8-30. That's not the same as getting to the wreck site. Nobody reached that.)
 

Paul Lee

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Hi Dave,
(What was the meaning of the Elgar note btw?)

The reason why I believe that Lord was right, more or less in his overnight position was because I cannot find any evidence, except for drift, that he was wrong. Alright, I have to assume that the 7.30pm starsight was right, giving a lattitude fo 42 5.5 N, but this is also consistent with his course during the remainder of the day. For his ship to get so far south (to be within visual range of the Titanic) or so far east (to coincide with my diagram) means that either (a) the drift was quite strong, and/or (b) the watchkeepers etc. on board the ship failed to notice that the ship was deviating drastically from its planned route.

Still, it doesn't explain the change in the Californian's ship bearings from SSE to SW. I just don't understand that.

Cheers

Paul

 

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