Celestial navigation practices


Georges G.

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George, More home work! Would have been a tragic loss sending Captain Sumner out the plank. He sure was no slouch. Any comments about which celestial objects other than the sun you use would be appreciated. Doug was kind enough to mention Betelgeuse.

Same question but what speed would you run with Olympic (1912) flat calm, no ice around however the ceiling has now lowered to sea level and the visibility is 1/4 mile in fog?

Now this question isn't really fair but given all you know about the Titanic and let say you were confident you wouldn't be in the vicinity of the ice until Murdoch's watch, what speed would you be running at?
Betelgeuse doesn’t pierce through an overcast sky any better than the sun! A sun sight can be taken say from 9:00 to 15:00 or during or during 6 hours. Betelgeuse can only be taken during the 25 minutes duration of the nautical dawn and during the 25 minutes duration of the nautical twilight or 50 minutes. So you have much more chance to find your position with the sun than a star. The moon is a very touchy one to use…

1912 Rules of the Road Article 16: Every vessel shall, in a fog, mist, falling snow, or heavy rainstorms, go at a moderate speed, having careful regard to the existing circumstances and conditions. A steam vessel hearing, apparently forward of her beam, the fog signal of a vessel the position of which is not ascertained, shall, so far as the circumstances of the case admit, stop her engines, and then navigate with caution until danger of collision is over.

The Elder Brethren, in advising Mr. Justice Bargrave Deane in The Counsellor, suggested that you ought not to go so fast in a fog that you cannot pull up within the distance that you can see, and his Lordship adopted this rule and said: If you cannot retain steerage way at such a speed then you should manage by alternately stopping and putting the engines ahead.

You had to be able to stop within the range of visibility. Out in the ocean, I think that Olympic under Haddock command would have proceeded in ¼ mile visibility at Standby Engines Half Ahead 50 RPM Turbine Engaged or 15½ knots. I am pretty sure that the twin screw vessel could have then be made dead stopped within two - three ship’s length at such speed with the engine crew standing by.

They also knew that a steam whistle audibility range in nautical was around 2 miles, giving thus a margin of safety. But in practice, the range at which a whistle may be heard is extremely variable and depends critically on weather conditions; the values given can be regarded as typical but under conditions of strong wind the range may be much reduced.

I have had the opportunity to meet thousands of ship’s masters. When things were smooth, I used to inquire ounce a while what their view on Titanic occurrence was. The vast majority responded that Smith must have been totally out of his mind to proceed at full sea speed or so, during a pitch dark moonless night while having been well advised of ice heavy ice ahead. Case closed!

Knowing myself as a prudent mariner, having a good experience in ice infested waters piloting, that the night was moonless pitch dark and having been well advised of ice ahead, I would not have proceeded faster than in ¼ mile visibility or at Standby Engines Half Ahead 50 RPM Turbine Engaged or 15½ knots from dawn to twilight. The time lost would’ve been totally insignificant and the vessel would have anyhow met her ETA.

One thing that I learned over time is that speed kills. Hitting an iceberg at 22½ knots makes exponentially much more damages that at 15½ knots. Kinetic Energy = ½mv2; therefore [15½ knots]² = 240 whereas [22½ knots]² = 500; thus the kinetic energy is multiply by a factor of [(500 – 240) ÷ 2] = 130 times !!! Consequently, glancing blow the berg at 15½ knots would not have opened 5 watertight compartments to sea.

Hypothetically, I think that I would have taken over Murdoch’s watch instead of going to bed. I can assure you that Boxhall would not have been in his cabin having a tea while on watch by the time the bell rang! I would have straighten the barque and the priorities; phoned down and order Standby Engines, stationed Murdoch on the starboard side, Boxhall on the port side, Moody by the quarter master, posted an extra lookout at the bow, supply glasses, end paper work. When the master is in charge on the bridge, everybody stands at attention. Realizing from my own eyes that it was a pitch dark moonless night, been well advised of ice ahead, knowing that the brand new vessel was carrying 2,200 souls at 22½ knots and not being too sure of the lifeboat capacity, my temptation to reduce to Half Ahead Turbine Engaged would’ve been very strong! Any ice sighted would have made me to reduce to Half Ahead Turbine Disengaged, and so on. When the master is fully in charge on the bridge … things have a real tendency to change!

Smith had much more shiphandling experience than Murdoch; who knows what he would have done or attempted …
 
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Mar 22, 2003
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The vast majority responded that Smith must have been totally out of his mind to proceed at full sea speed or so, during a pitch dark moonless night while having been well advised of ice heavy ice ahead. Case closed!
Bear in mind that back in 1912 it was assumed that an iceberg could be spotted as far out as 2 miles on a clear dark, moonless night. We now know that that assumption is fallacious. About 1/2 mile is about right for a medium sized berg, and less than that for small bergs and growlers. I do agree with the measures you stated since Smith didn't take his ship further south before turning for NY.
Hitting an iceberg at 22½ knots makes exponentially much more damages that at 15½ knots.
The kinetic energy at 22.5 knots compared to 15.5 knots is a factor of 2.1 times greater; (22.5/15.5)² = 2.1. Still, it would have been significantly less damaging at 15.5 knots.
 

David Allison

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Smith had much more shiphandling experience than Murdoch; who knows what he would have done or attempted …
George, That is one excellent post. Thank you for the comments and taking the time to write it out. Give's me much to think about. I due have some comments, however due to a pending medical procedure, won't be able to get back to you for a few days,

Dave

PS - Rare photo attached taken of Captain Sumner interning while attending Harvard :)
 

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Georges G.

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The kinetic energy at 22.5 knots compared to 15.5 knots is a factor of 2.1 times greater; (22.5/15.5)² = 2.1. Still, it would have been significantly less damaging at 15.5 knots.
It is not what I wanted to say? ;)

Thanks Samuel for the correction ... :)
 
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David Allison

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George,

This may be steering the topic of the thread seriously off course. Perhaps “Safety, Speed and Senior Officers” of which celestial navigation is one part? Just wondering if perhaps a new thread should be started as I have quite a few comments of which I’d like your comment.

Intuition tells me there is something inherent in your ability as a professional mariner (Ship handler/Pilot) which you may not be aware of. I believe for example, in a hypothetical situation if you were transported back in time and we modified the bridge of Titanic (moved the wheel) to a position where you were the senior officer of the watch, with helm control firmly in hand, I believe you would have handily avoided hitting the iceberg. Alternatively someone less experienced, the result would remain unchanged. As to why I believe this is not particularly easy to explain. Perhaps if I first start with a Captain Sumner analogy this may help? It’s your bath time, George!

Captain Sumner has an acquaintance named Bob who owns a box barge of length (L) and width (W). Having studied integral calculus at Harvard, Captain Sumner performs a mathematical calculation on Bob’s box barge, which yields the result L/6. Now Captain Sumner is puzzled by this result for the reason there is no W (Width). The result is only dependant on the Length of the barge, not its width. As time passes, Captain Sumner struggles to understand this and searches for an observation confirming his findings. Incessantly curious he comes up with a plan.

Its Captain Sumner’s bath time and the boys have prepared his bath. Captain Sumner decides to float several rectangular pieces of wood, each varying in length and width, in his bath tub for testing. One piece of wood has a higher density. They all share the same thickness.

For each piece of wood he has drawn a dotted line down its center line, dividing each box barge’s length into 6 equal parts, according to his mathematical result L/6. Below is one such barge.

Captain Sumner jumps into the tub and proceeds to poke each piece of wood at the location and direction of the green arrow in the sketch. Poking is applying a force sometimes referred to as moment. Captain Sumner then observes where rotation first occurs.

Bob.JPG


Captain Sumner makes the following observations at time = 0 when he first pokes the wood.
  1. If he applies the moment quickly, rotation first occurs about point ‘B’
  2. Regardless of how quickly he applies this moment, he is unable to move the point of rotation ahead of point ‘B’ towards the bow.
  3. If he applies the moment very slowly he is almost able to have the barge rotate about point ‘C’ (L/2) it’s center of mass.
  4. The density of wood does not seem to affect these results. As long as the wood floats, these reactions remain consistent with each different piece of wood.
Captain Sumner decides to call point B the primary pivot point of all box barges located L/6 distance from the box barge’s center of mass which is L/3 from its bow.

Now Captain Sumner removes himself from the tub. Standing shivering holding a towel he shouts “DRAIN THE TUB!” His men dutifully start draining the tub. When the tub is almost empty Captain Sumner shouts “STOP”. At this point the barge with the highest density is grounded sitting on the bottom of the tub, which is flat. He now repeats his test on the remaining barges which are floating.

Bob 2.JPG


Captain Sumner makes the following observations at time = 0 when he first pokes the wood.
  1. If he applies the moment quickly, rotation first occurs about point ‘A’ which is L/3 from point C, the box barge’s center of mass.
  2. Regardless of how quickly he applies the moment, he is unable to move the point of rotation ahead of point ‘A’ towards the bow.
  3. If he applies the moment very very slowly he is almost able to get the box barge to rotate about point ‘C’ its center of mass.
  4. As long as the wood floats, these reactions remain consistent with each different barge.
Captain Sumner decides to call point A the secondary pivot point of the barge located L/3 distance from the box barge’s center of mass, L/6 from its bow.

Length AC = 2 x BC

Now Captain Sumner believes this information important to share with Bob. Bob on the other hand fancies himself an expert in pretty much everything related to box barges. He says to Captain Sumner “Son, I was piloting box barges before you were born.” After several minutes in a rant Bob concludes by saying,

“Now I’ve finished talking about me being an expert on the subject of box barges. Now I would like to hear you talk about me being an expert on the subject of box barges.”

Captain Sumner excuses himself and returns to his crew.
 

Georges G.

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Hello David,

You seem to discuss about two different matters; one being the «Bridge Parallax» and the other, the «Center of Lateral Resistance vs. Apparent Pivot Point vs. Interaction».

I am not too sure about which one you are interested with?
 

David Allison

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Hi George,

These points which total 5 and divide the length of the box barge into 6 equal lengths, are a function of the geometric shape of box barges displaced volume which in turn is related to the mass of the ship. I didn't want to give the impression the points have anything to do with Bridge Parallax.

The displaced volume is a very simple shape for a box barge which has a length, width and depth. Changing the width of the barge does not affect the location of these points, located along the ships longitudinal axis. The primary and secondary pivot points are not apparent, but rather precise points which remain fixed. Points which are invisible to us.

These points play a key role in determining the minimum turning circle radius for the box barge, for example after we set the barge in motion and attempt a turn.

I believe its common knowledge ships operating in shallow waters risk doubling their minimum turning circle radius at slow speeds. At least I hope this is common knowledge. This provides a reasonable explanation of why this occurs. Everyone can demonstrate for themselves these points exist. I don't see any advantage of posting the mathematics involved. We can associate the primary and secondary pivot points with a ship's speed. In deep water at higher speeds the ship uses the secondary pivot point, closer to the bow when a turn is attempted. Of course the box barge has drag issues.

In the example we are supplying the moment. In practice the ship provides the moment. In a hard over rudder manoeuvre this moment (force) starts at zero, reaches some maximum and returns to zero as the rudder reaches full deflection. This force is huge for large displacement vessels. Perhaps I jumped the gun on this and went right to one of my conclusions. This being an experienced ship handler unconsciously controls this moment by the rate he/she turns the wheel. The word swing comes to mind. He/she becomes accustomed not only to how much input is needed to obtain a result, but also to the significant lead and lag times involved.

Please understand I may not have any idea what I'm talking about. Send me out the plank at will. For me this was the first step of many which has lead me to certain conclusions. I was hoping to engage someone experienced in ship handling like yourself to discuss this further. (email is fine in you prefer, or if you prefer not, I understand) The conversation will involve inertia of large displacement ships which I believe is a topic not well understood.

Have I answered all of your questions?

Thanks George, btw it allison not allision ;)
 

Georges G.

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«Allison not Allision» … obviously you were predestined! ;)

David, what is your ultimate shiphandling question; as straight forward and as short as possible?
 

David Allison

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Perhaps its not so much a question George but a desire to set the record straight. I doubt for example people know the harder you push ships like Titanic and Olympic, the shorter their stopping distance. I doubt they know for example at half speed they significantly increase their stopping distance, compared with the full speed stopping distance.
David, what is your ultimate shiphandling question; as straight forward and as short as possible?
To avoid a large obstacle directly one ship length ahead, in a ship like the Olympic or Titanic proceeding at 22 knots, would you spin the wheel as rapidly as possible to its limits, to avoid this obstacle?
 
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I doubt they know for example at half speed they significantly increase their stopping distance, compared with the full speed stopping distance.
Huh? Do you mean a ship going at half speed ahead will travel a longer distance before stopping than a ship going full speed ahead will from the point when the engines are stopped?
would you spin the wheel as rapidly as possible to its limits, to avoid this obstacle?
You can spin the wheel as fast as you can but in those ships its the speed of the steering engine that will determine how long it takes for the rudder to go full over. On these ships full over was 40°. (Most vessels full over is about 35°.) And, by the way, the force on the rudder does not drop to zero when full over. It reaches a peak close to its stall point.
 
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David Allison

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Sam, I'll respond to you and Doug, however I'd first like George to comment on what I said. Then give me some time after George's comments. dave
 

Georges G.

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The precautions expected by the ordinary maritime practice makes a seaman facing an immediate and inevitable peril at sea, not liable for a given action if the expected outcome is not the one obtained. As well, one cannot be accountable for an anticipated result without having the related authority and resources to achieve it. Nobody is bound to be a hero or to do the impossible.

Have a look to what is being teach these days; «When collision with another vessel is considered to be inevitable, the foremost concern of the officer must be to maneuver his ship so as to reduce the effect of collision as much as possible. The consequences are likely to be most serious if one vessel strikes the other at a large angle near the mid length. The engines and the helm should be used so as to achieve a glancing blow rather than a direct impact. The damage would probably be the least serious if the impact is taken forward of the collision bulkhead.» CAPTAIN ALFRED NORMAN COCKCROFT Extra Master, M.Phil., FRIN, FNI.

Therefore, I really don’t see what could have been done better than what was done by Mate Murdoch. The more you know about that split minute occurrence the less turn out the available options.

Note: The avoiding manoeuver has been simulated over and over again by certified shiphandlers instructors. As far as I know, no one ever succeeded to avoid such a close by berg at such a speed. If someone’s attempt had succeeded, don’t worry we would have heard about it. However at half head, the maneuver succeeds…
 
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