How Far South did the Titanic Reach?

Mar 22, 2003
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the only thing square in my time was a piece of a chocolate bar:
Hmmm, the only chocolate bars I ever had were rectangular.:p (As an aside, in the plot in my previous post, I extended the intercept line in the direction of the celestial body with an arrow ending with the symbol of the celestial body. In this instance the sun. That practice was fading into disuse, but some yachtsmen liked doing it on their chart paper. The EP was a point on the LOP that was at the shortest distance between the DR and the LOP line. Probably that's why a little square was used because of the right angle formed between the intercept and LOP lines? Just a guess. Other than that, the EP has little meaning. This was the practice used back in the 1970s.)

Anyway, I was taught that the course made good can only be determined after a second fix is obtained, after which you can then get the set and drift of the total average current that affected the ship by comparing the DR of the fix to the actual fix itself. Yes, if know the current you can compensate for it by adjusting your course for it. Did that all the time in my flying days using predicted winds aloft.
Anyway, the log doesn't measure current, only distance travelled through the water. If the ship was making say 22 knots at 75rpm through the water, and if there was a head current of 1 knot, the log would still measure an advance of 22 miles each hour assuming it was accurate. And let's not forget that Boxhall said that he thought the ship would reach the corner some "considerable time" before 5:50pm. To me there had to be some indication to him that the vessel was running ahead of her DR, not behind it, after he came on duty at 4pm. Unless, of course, you want to believe that he was outright lying about that?
 

Jim Currie

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The course made good is most certainly the CMG between two fixes but if you have one or more alterations of course between these two fixes then that definition becomes too precise. In my sketch, the 2nd CMG is implied.
If you want to keep to the world of Captain Smith... then he laid off the current for 4 hours from the 4 pm DR...determined where he thought his ship was at that time. He then calculated the shortest distance and course to steer to bring his ship to a point whereby he could turn onto the next course. My 2nd CMG illustrates what I believe happened, give or take a mile.

The Patent Log was simply an odometer which recorded the number of miles the instrument traveled in a given time. Since it was attached to a ship, it also measured the number of miles the ship had traveled in the same time. By your reckoning, it measured speed, not distance. Think about it!
If , as you say, it recorded a distance of 22 miles every hour then at 6 pm, it would have recorded a total distance of 6 x 22 = 132 miles and 5th officer Lowe would gave told Senator Smith:

"6241. Yes. - Her speed from noon until we turned the corner was just a fraction under 22.1 knots... If you take the average speed from 12 to 6 - that is giving her a run of six hours....
6257. But you had means, had you not, of ascertaining definitely how fast the ship was going? A: - In what way, sir? We have the log -
6258. (interposing). Between 6 and 8 o'clock. A: - We have the log.


In fact, Lowe told Senator Smith:

"it was really 20.95, about. If the speed had been increased or reduced during the interval when I was off duty, I would have been informed of it".

Why do you keep burying your head in the sand over this evidence, Sam?



 
Mar 12, 2011
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The patent log is just a little device with a propeller on it connected to a cable, which records the revolutions of the propeller and displays them in terms of nautical miles on a dial on the stern of the ship, correct? Much like how the odometer of a car measures distance traveled by revolutions of the tires. If I'm completely mistaken, none of the following is going to make sense so I apologize in advance. The propeller in the patent log would spin at the same speed if the ship was making 22 knots through the water, whether she was being acted on by a current or not. Just like the ships propellers don't slow down when the ship has a current on her head, neither would the patent log.

Much like how an odometer can be thrown off by outside factors (using a larger or smaller size of tire than it was calibrated for, as an example), so can the patent log be thrown off by factors that affect the distance the ship travels over the ground. So if the ship travels at 22 knots for an hour, but is acted on by a 1 knot current running on the exact reciprocal of her course (unlikely, but bear with me for the sake of argument) she will only make good 21 miles over the ground. The patent log will still say 22 miles. I think it's an oversimplification to say it measures speed, rather it measures the distance the ship covered before extenuating factors such as current and wind are factored in. So at 6pm, the log WOULD have shown 132 miles traveled, but the actual distance traveled would have been something less if the ship was travelling at 22 knots through the water.

I've done some of my own math working at this problem myself, and I realized my methodology is flawed because I've been using flattened sphere calculations rather than the round sphere calculations that everyone else is using. I will have to do some more work and post my results later.
 
Nov 26, 2016
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So at 6pm, the log WOULD have shown 132 miles traveled, but the actual distance traveled would have been something less if the ship was travelling at 22 knots through the water.
The log measures miles travelled through the water.
Case 1, no wind, no current, log reading 132 miles in six hours, 132 miles made good, 22 knots over ground.
Case 2, no wind, 1 knot head current, log reading 132 miles in six hours, 126 miles made good, 21 knots over ground.
Case 3, westerly wind, no head current, log reading 126 miles in six hours, 126 miles made good, 21 knots over ground.
 
Mar 12, 2011
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Ah, you're right. Wind would affect speed through the water, current would not. I wasn't sure. I'll edit my post accordingly. (Never mind, I guess there's a time limit on editing posts here.)
 

Jim Currie

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Hello Lads.

This discussion is interesting but academic. The fact of the matter is that if Titanic's 5th Officer Lowe used a precise speed of 20.95 knots to work a DR position, he did not pluck that number from the sky. He must have got it from a Patent Log Reading. He came on Watch at 6 pm that evening. At that time, the Patent Log reading would have been passed to the bridge. He would simply have divided that reading by 6 to get an average speed from Noon that day. Now ask yourselves this:
Why is that evidence rejected yet the 10 pm Patent Log reading given by QM Hichens in his evidence showing a speed of 22.5 knots is universally accepted without question?

As for the Patent Log? I've streamed and used one of these things more times than I can remember. However what I do remember is that despite the weather conditions, including fierce storms, they recorded the distance traveled with an accuracy that was only restricted by mechanical failure, improper streaming, initial calibration or fouling by floating objects. Here is an extract from examination questions and answers posed to masters and Mates at the time of, and after the Titanic disaster, The red emphasis is mine:
Patent Logs. 001.jpg
 
Mar 22, 2003
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This discussion is interesting but academic.
Yes it is because many assumptions are being made. Case in point:
He must have got it from a Patent Log Reading. He came on Watch at 6 pm that evening. At that time, the Patent Log reading would have been passed to the bridge. He would simply have divided that reading by 6 to get an average speed from Noon that day.
Lowe actually said that he got that speed by: "I used the speed for the position at 8 o'clock, and got it by dividing the distance from noon to the corner by the time that had elapsed from noon until the time we were at the corner." It was also Lowe who worked the course and distance from noon to the corner while he was on duty from noon till 4pm, "I worked the course from noon until what we call the 'corner'; that is, 42 north, 47 west. I really forget the course now. It is 60º 33 1/2' west - that is as near as I can remember - and 162 [126?] miles to the corner." When he spoke of the use of the log it was in response to being asked how he knew for certain that he used the right speed in getting his 8pm DR. Lowe's response was: "In what way, sir? We have the log -" He was then asked by Sen. Smith, "And inasmuch as you did not take the revolutions, I wondered whether you were strictly accurate when you defined the ship's position at 8 o'clock." Lowe then went on to talk about the unfinished slip table they were working on, and mentioned that if they had increased or decreased the speed while he was off duty (between 4 and 6pm) then he would have been informed of it.

With respect to the log and revolutions, Lowe also pointed out, "We ring him up, and we see how she is doing with the revolutions, whether she is going faster or going slower; and you will find a corresponding difference in the log." If the log registered 126 miles in 6 hours, then it would register an advance of 42 miles in two hours with the same number of revolutions. Yet we were told that the log advanced about 45 miles in two hours. That's an increase of 3 miles in two hours, or just over 7%. For that to happen, the revolutions would have to have increased from 75rpm to about 80rpm, which did not happen. The other data point in log miles came from Rowe who took the reading when the struck and said it was 260 miles. If the time was 11h and 40m, then that is an average advance of 44.6 miles every two hours, which is certainly in keeping with 75rpm. (If the actual advance between 8 and 10pm was 45.0 miles, then that would imply the revolutions over that two hour period had increased by 0.7 rpm from the 75rpm average. But my guess was that 45 miles was a rounded number to begin with.)
Case 3, westerly wind, no head current, log reading 126 miles in six hours, 126 miles made good, 21 knots over ground.
It would have to be a relatively strong head wind to loose a full knot in speed. Besides, the moderate wind that afternoon came over the ship's starboard quarter.
 

Jim Currie

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Yes it is because many assumptions are being made. Case in point:

Lowe actually said that he got that speed by: "I used the speed for the position at 8 o'clock, and got it by dividing the distance from noon to the corner by the time that had elapsed from noon until the time we were at the corner." It was also Lowe who worked the course and distance from noon to the corner while he was on duty from noon till 4pm, "I worked the course from noon until what we call the 'corner'; that is, 42 north, 47 west. I really forget the course now. It is 60º 33 1/2' west - that is as near as I can remember - and 162 [126?] miles to the corner." When he spoke of the use of the log it was in response to being asked how he knew for certain that he used the right speed in getting his 8pm DR. Lowe's response was: "In what way, sir? We have the log -" He was then asked by Sen. Smith, "And inasmuch as you did not take the revolutions, I wondered whether you were strictly accurate when you defined the ship's position at 8 o'clock." Lowe then went on to talk about the unfinished slip table they were working on, and mentioned that if they had increased or decreased the speed while he was off duty (between 4 and 6pm) then he would have been informed of it.

With respect to the log and revolutions, Lowe also pointed out, "We ring him up, and we see how she is doing with the revolutions, whether she is going faster or going slower; and you will find a corresponding difference in the log." If the log registered 126 miles in 6 hours, then it would register an advance of 42 miles in two hours with the same number of revolutions. Yet we were told that the log advanced about 45 miles in two hours. That's an increase of 3 miles in two hours, or just over 7%. For that to happen, the revolutions would have to have increased from 75rpm to about 80rpm, which did not happen. The other data point in log miles came from Rowe who took the reading when the struck and said it was 260 miles. If the time was 11h and 40m, then that is an average advance of 44.6 miles every two hours, which is certainly in keeping with 75rpm. (If the actual advance between 8 and 10pm was 45.0 miles, then that would imply the revolutions over that two hour period had increased by 0.7 rpm from the 75rpm average. But my guess was that 45 miles was a rounded number to begin with.)

It would have to be a relatively strong head wind to loose a full knot in speed. Besides, the moderate wind that afternoon came over the ship's starboard quarter.
That will not do Sam. If you wish to go down the road to a satisfactory conclusion to this debate, you must resist the temptation to treat the evidence like a biased journalist or rather; the Editor of a biased newspaper.

Third Officer Lowe was giving evidence at a hearing to discover what happened during the worst peacetime maritime disaster in history. A disaster which he survived. He was doing so 8 days after the event...8 days filled with constant questions on a myriad of subjects. You cannot expect him or any normal young person to have been A1 perfect with his recollections of tasks which at the time they were performed, he had no reason to store in his memory in case they were asked for at a later date. That's nonsense.

I suggest you outright declare Lowe's evidence to be pure fantasy. However, I'll "take the bait" ... can't resist it.

If you can't trash Lowe's speed then you must accept that he did not pluck it from the air. In the evidence you quoted Lowe, he said he got 20.95 knots by dividing the distance run to the turn by 5 hours 50 minutes. If so, then the distance run was 122.2 miles when Titanic turned. However, he was not on the bridge at 5-50 pm so any speed he used must have been obtained after that time. He would only have been able to determine an average speed at or after 6 pm when he came on Watch.
Are you suggesting that Lowe mentioned the word "Log" simply to remind the good Senator that the ship was fitted with one? That's absurd, Sam. In fact, if we are to believe any of the evidence given by Lowe or his boss, Pitman, they used the log quite a bit during their Watch.

As for the sudden increase in speed:

If a ship makes 22 knots at 75 rpm under normal conditions and suddenly encounters a 1 knot head current, her speed will drop by about 1 knot. If that current is suddenly removed and at the same time all other external influences are removed, the ship's speed will increase almost immediately to the optimum for the prevailing conditions. That's what happened in the case of Titanic. That's why Boxhall used 22 knots instead of 21 knots. if he had used the Patent Log instead of rpm the outcome would have been a little different...he would have been even farther ahead with his distress DR,

A ship's speed is very sensitive to many influences. Principal of these are wind, sea & swell and current. Remove all vestiges of these suddenly and suddenly, the speed will jump up. Not only that, but suddenly, unheard vibration sounds become amplified.
The obstructions to he ship's progress disappeared with the sun. That, believe it or not, is a very common occurrence at sea.
 
Mar 22, 2003
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You still seem to be mixing distance by log with distance made good.

From the daily mileage runs we were told, the vessel travelled 1549 miles to noon on the 14th. That leaves about 126 miles to the corner, consistent with a total travel distance from Daunts Rock LV to the corner of 1675 miles. (From Olympic data from 4 voyages commanded by EJ Smith over the same route of travel as Titanic, we find distances from Daunts Rock to the corner of: V1 1677 miles, V2 1674 miles,V3 1676 miles, and V8 1675 miles.) Lowe even said he calculated the ships course from noon to the corner that day. It's no secret that he would have known the ship was 126 miles before the corner at noon. Based on that little chit of paper he handed to Sen. Smith, and Smith's immediate response, what he showed Smith is how he got a speed of 21 knots by simply dividing 126 miles by the 6 hours that he mentioned. He then said it was really a little less than that, 20.95 knots, which would make the distance 125.7 miles if he cared to be more exact.

If we go with your belief that the log measured 122.2 miles at 5:50pm, then there is no way that Boxhall, or anyone else for that matter, could claim that the ship was to the south and west of the corner when she turned at 5:50pm. A run of 126 miles in six hours (or 122.2 in 5h 50m) yielding a speed through the water of 21 knots is inconsistent with an advance of nearly 45 miles in two hours seen at 10pm. In fact, it is inconsistent with Olympic making 21.5 knots carrying 74 revolutions which comes from Wilding. And Titanic was expected to make about 1/4 knot better than her sister ship according to an admission from Ismay. (At 75 revolutions Olympic would have ran about 21.8 knots.) And since the log "indicates the distance travelled through the water, not the distance made good," a run of only 126 miles by log in 6 hours, or 21 knots, is certainly inconsistent with Titanic carrying 75rpm; unless, of course, you want suggest that her revolutions had gone down to about 71 or 72rpm over that six hour period which goes against what was reported.
 
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about wind and speed
Olympic run data february 1912, day 2,3,4
feb 10, 536 miles, 21,57 knots, fresh WNW wind to fresh N'ly gale, high swell
feb 11, 545 miles, 22,02 knots, mod. N to light var. to mod. WNW wind
feb 12, 563 miles, 22,70 knots, mod. N to mod. NE gale, rough NNE sea

Beesley wrote about the weather on Sunday April 14:
There is not much to tell after leaving Queenstown from Thursday to Sunday. The sea was calm, in fact as calm that only a few passengers omitted their meals.The wind blowing from west or south west, "fresh" as the daily weather chart indicated ...

About Sunday: ... when we went on deck after lunch we noticed a change of temperature of that kind, that not many preferred to expose themselves to the cold wind - an artificial wind which mainly if not completely was caused by the fast run of the ship. ... I am sure there was no wind wawing at this time. I noticed the same strength of wind arriving at Queenstown, which died off when we stopped, and revived immediately when we left the Harbour ...

It seems there is no wind available to explain a slow down from 22 to 21 knots sunday afternoon.
 
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I've done some of my own math working at this problem myself, and I realized my methodology is flawed because I've been using flattened sphere calculations rather than the round sphere calculations that everyone else is using. I will have to do some more work and post my results later.
I think the differences between flattened and round sphere calculations can be neglected.
The earth's radius is 6357 Kilometers near the pole and 6378 kilometers at the equator. Travelling by 1' of longitude with one or the other radius would make either 1849 or 1855 meters. The official definiton of 1 nautical mile is 1852 meters. The error would be just 0,16%, even less as we are travelling between 40 and 50 degrees latitude. Another formula for Bessel ellipsoid says:

1' = 1852 - 9,3 * cos (2 * degree latitude)
This makes 1853,6 meters at 50° and 1850,4 meters at 40° latitude.
1,6 meters of 1852 meters, percentage 0,086%.
Supposed the log is calibrated to register "1852 meter" miles, the indication would be 132,11 instead of 132 miles.

The accuracy of stellar fixes is about one mile, so the error caused by not accounting for the flattened sphere is neglectable related to the normal inaccuracy of stellar navigation.
 
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Markus
You're absolutely right. I was doubting my figures because I was coming up with positions slightly different from Sam Halpern and other expert navigators here (which I am not) but the differences are so small that it doesn't make much difference in the scheme of things. I'm teaching myself the math as I go along, they don't make network engineers take math quite this complex these days!
 

Jim Currie

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You still seem to be mixing distance by log with distance made good.

From the daily mileage runs we were told, the vessel travelled 1549 miles to noon on the 14th. That leaves about 126 miles to the corner, consistent with a total travel distance from Daunts Rock LV to the corner of 1675 miles. (From Olympic data from 4 voyages commanded by EJ Smith over the same route of travel as Titanic, we find distances from Daunts Rock to the corner of: V1 1677 miles, V2 1674 miles,V3 1676 miles, and V8 1675 miles.) Lowe even said he calculated the ships course from noon to the corner that day. It's no secret that he would have known the ship was 126 miles before the corner at noon. Based on that little chit of paper he handed to Sen. Smith, and Smith's immediate response, what he showed Smith is how he got a speed of 21 knots by simply dividing 126 miles by the 6 hours that he mentioned. He then said it was really a little less than that, 20.95 knots, which would make the distance 125.7 miles if he cared to be more exact.

If we go with your belief that the log measured 122.2 miles at 5:50pm, then there is no way that Boxhall, or anyone else for that matter, could claim that the ship was to the south and west of the corner when she turned at 5:50pm. A run of 126 miles in six hours (or 122.2 in 5h 50m) yielding a speed through the water of 21 knots is inconsistent with an advance of nearly 45 miles in two hours seen at 10pm. In fact, it is inconsistent with Olympic making 21.5 knots carrying 74 revolutions which comes from Wilding. And Titanic was expected to make about 1/4 knot better than her sister ship according to an admission from Ismay. (At 75 revolutions Olympic would have ran about 21.8 knots.) And since the log "indicates the distance travelled through the water, not the distance made good," a run of only 126 miles by log in 6 hours, or 21 knots, is certainly inconsistent with Titanic carrying 75rpm; unless, of course, you want suggest that her revolutions had gone down to about 71 or 72rpm over that six hour period which goes against what was reported.
Not in the least 'mixed-up', Sam.

No problem with Lowe's thought processes either. I don't 'believe' that the log showed 122.2 miles at 5-50 pm; it must have done so if Lowe got a speed of 20.95 knots via a log reading for 6 pm.

There would not be a single officer on the bridge of Titanic who did not have a number in mind as to the distance to go from Noon to The Corner. All of them would have his own idea of what speed the ship would make over that distance.

Boxhall freely admitted that he did not rely on the patent log but used engine rpm. Whereas, Pitman and Lowe both stated that they used the patent log. It's all in the evidence, Sam.
One thing none of them, including Captain Smith would know for sure, was the conditions of wind and weather to be encountered.
You can be sure that Captain Smith would be on the lookout for his ship slowing down. It was and still is, a distinct possibility event in that area. However, his planned clock change of 47 minutes tells us that he hoped that Titanic would cover about 540.5 miles between Noon April 14 and Noon April 15.at the same rpm. That's 5.5 miles less than the previous day. so he was expecting to slow down at some point.

Consider this: if there was a partial clock change of 24 minutes then when Titanic hit the iceberg she would had been running for 12 hours and 4 minutes and covered a distance of 260 miles. At an average speed of 22 knots and under normal conditions; for the next 12 hours and 23 minutes, Titanic would have covered a distance of 272.4 miles, making a total of 532.4 miles...8 miles...less that Captain Smith anticipated and 13.6 miles less than the previous day's run at exactly the same rpm. During Noon 14/15, Titanic was running between 44-33'West and 56-18'West.
Is it shear coincidence that 10 months earlier, in June, 1911 and during her maiden voyage,Titanic's sister ship Olympic lost 17 miles of her expected speed while running between 43-52'West and 54-47'West?
On both occasions both vessels passed along exactly the same route in exactly the same area. Check it out for yourself.
Incidentally, has Titanic survived, and encountered SW'ly weather over the second half of the run after Midnight as did Olympic, she would have most certainly have lost more than 13.6 miles.... perhaps event that extra 3.4 miles?

Since Boxhall and Pitman believed Titanic had turned late, we can safely assume that they believed she had been allowed to run past The Corner. If both these men used the same speed and distance from Noon (I'm suspicious of Pitman) then both would have believed the ship had turned at the same time and consequently, the same distance beyond the planned turning point. They did not.
Pitman waffled-on about a turning time of 5 pm which was ludicrous, in particular if combined which his over-shoot of 10 miles. However therein may lie the cause of the mysterious extra 20 miles for the 8pm DR?
Boxhall obviously thought the ship had over-shot. Otherwise how could Titanic have been 'right on the track' at 7-30pm sights and make good a course up to then of 266 True? If anything, she should have been gradually making a course to the southward of the intended track as the local Magnetic Variation reduced.
If there had been no clock adjustment before impact, Titanic would have been at about 41-56.3'North, 47-56'West at 7-38 pm that evening. If, as claimed, she was right on the track at that moment then she was 42 miles on a bearing of 265 True from The Corner. If, as Boxhall claimed, she made 266 True from 5-50 pm when she turned, then she covered a distance of 38.2 miles at a speed of 21.5 knots...not 22 knots and was at about 41-59'North, 47-05'West.
On the other hand, had the clock been retarded 24 minutes before impact, then Titanic would have been 33 miles x 265 True from The Corner at 7-38 pm. If she had averaged 21.5 knots from when she turned, at 5-50 pm, she would have turned 5.8 miles on a bearing of 065 True from The Corner when she was at 41-59.9#North, 46=51'West.
Think about it! Boxhall 'fitted' his DR 5-50 pm turn position to the 7-30 pm fix position.
Not only did he use the wrong speed and time, but he also used the wrong course when calculating his distress position. here's where the 'fit' comes in:

15671. The effect would be she would have run a little bit further on the old course and then on the new course she is gradually making back to the line?
- That is my impression of the idea which Captain Smith had in altering that course and setting it to that time."

The he states:

"4976. How many degrees did you change ? A: - I can not remember. If I had a chart here I could tell you in a minute. South 84 or 86 west would be the true course we were making after 5.50; south 84 or 86, I am not quite certain which, was the true course."

T
he man was under pressure and waffling, Sam.

 
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Mar 22, 2003
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Michael,

I've been using what is called the mid-latitude method in my spreadsheet. Given a starting point with coordinates Lat1, Lon1, for vessel traveling D nautical miles on a course C, you can use the following equations to get change in latitude, dLat, and the change in longitude, dLon, both in minutes-of-arc to add or subtract from Lat1 and Lon1 to get to the ending point coordinates at Lat2, Lon2.

dLat = D cos C in miles
dLon = D sin C / cos Lmid

where: Lmid = (Lat1 + Lat2)/2

Lat2 = Lat1 + dLat
Lon2 = Lon1 - dLon

Step-by-step example: Lat1 = 42° 00' N, Lon1 = 47° 00'W. Ship traveling 22 knots for 18h and 57m on course 266° true. Find Lat2, Lon2.

First we find distance travelled, D = 22 x (18h 57m) = 22 x 18.95 = 416.9 miles.

Then dLat = 416.9 cos 266° = 416.9 x -0.0698 = -29.1 miles = -29.1' [negative indicating southward movement]

Then we find Lat2 = Lat1 + dLat = 42° 00' - 29.1' = 41° 60' - 29.1 = 41° 30.9' N.

Next we find Lmid = (Lat1 + Lat2)/2 = (42° 00' + 41° 30.9')/2 = (42.00° + 41.52°)/2 = 83.52°/2 = 41.76°

Now we can find dLon in minutes-of-arc = D sin C / cos Lmid = 416.9 sin 266° / cos 41.76° = - 416.9 x 0.9976 / 0.7459= -557.5' [negative here means westward]

Now we can get Lon2 = Lon1 - dLon = 47° 00' - (-557.5') = 47° + 557.5' = 47° + 9° 17.5' = 56° 17.5' W.

Thus we end up at 41° 30.9' N, 56° 17.5' W.
 
Mar 22, 2003
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can not remember. If I had a chart here I could.....
Jim, that was Pitman not Boxhall. I agree that Pitman's 5pm at the corner was ludicrous, but he got that by reversing what he knew by then. He took for granted that the ship reached Boxhall's CQD location at 11:46pm. The CQD is 145 miles from the corner. At Pitman's 21.5 knots it would take 6h 45m to get back to the corner. 11:46-6:45 = 5:01pm.
Simple reasoning, but wrong.
 
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However, his planned clock change of 47 minutes tells us that he hoped that Titanic would cover about 540.5 miles between Noon April 14 and Noon April 15.at the same rpm. That's 5.5 miles less than the previous day. so he was expecting to slow down at some point.
Jim, I think one can not conclude from 47 minutes that a slow down was exspected. The clocks are changed by integer number of minutes and the 47 minutes are the result of round up or down. I think Smith calculated that way:

22 knots, 24,8 hours, 545,6 miles to exspect.
Starting Point: 43-02 N 44-31 W
Monday noon: 41-23 N 56-20 W
Difference of longitude: 11°49' * 4 = 47 minutes 16 seconds, round down 47 minutes.

Other possibility:
21,5 knots, 24,7 hours, 532,1 miles to exspect
Starting Point: 43-02 N 44-31 W
Monday noon: 41-24 N 56-02 W
Difference of longitude: 11°31' * 4 = 46 minutes 4 seconds, round down 46 minutes.

1 Minute of time represents 11 miles more or less.
The Input value for result 47 minutes will be 543 miles plus/minus 5.
They intended to alter the clock by 47 minutes, so most likely they exspected 22 knots for the next day.

Boxhall obviously thought the ship had over-shot. Otherwise how could Titanic have been 'right on the track' at 7-30pm sights and make good a course up to then of 266 True? If anything, she should have been gradually making a course to the southward of the intended track as the local Magnetic Variation reduced.
What is the tolerance zone for "right on the track"? You are native speakers, I am not. But based on my stomach feelings I should say if the star position is found one or three miles south of the track after a run of 170 miles after the last fix this is a fairly good result, the term "right on the track" still may be used.

Boxhall's CQD has been calculated either with course 265 or 266. It is located one mile south of the track. That means, the 7-30-or-40 position must have been 1 mile south of the track if calculated with 265°, or 2.5 miles south of the track if calculated with 266°.
This is just the first uncertainty. The second uncertainty arises by the error of Boxhall's CQD. If only he made a speed or time error, the correct CQD would be one mile south of the track, but if he switched the 42/48 columns in the traverse tables the latitude would be the same, the longitude would be 50-00, the correct CQD would be 2 miles south of the track. Smith's wrong CQD is 2 miles south of the track! At least these two would coroborate then!
Thus we have four possibilities to relocate the 7-30-or-40 Position:
1 / 2.5 / 2 / 3.5 miles south of the track.

The 265/266 uncertainty:
Pitman said in USA "South 84 or 86 west would be the true course we were making after 5.50; south 84 or 86, I am not quite certain which, was the true Course...
May be we can take this as evidence for 266. He knew they were 1 degree deviating from the prescribed course, he just could not remember which side. What bothers me however is Boxhall:
15671. The effect would be she would have run a little bit further on the old course and then on the new course she is gradually making back to the line?
- That is my impression of the idea which Captain Smith had in altering that course and setting it to that time."

He explained immediately before that 265 was the prescribed course, and 266 instead was steered to compensate the late turn.
This was his "Impression".
But was this "Impression" based on a calculation "afterwards" to find a turning Point at 5.50 which would match his wrong CQD Position?

Such turning point could have been:
41-55:30 N 47-10:30 West
Distance to 41-46 N 50-14 W: 137 miles; Course 86°
Speed with/without clock retarded: 21,3 / 23,5 knots
 
Mar 12, 2011
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Michael,

I've been using what is called the mid-latitude method in my spreadsheet. Given a starting point with coordinates Lat1, Lon1, for vessel traveling D nautical miles on a course C, you can use the following equations to get change in latitude, dLat, and the change in longitude, dLon, both in minutes-of-arc to add or subtract from Lat1 and Lon1 to get to the ending point coordinates at Lat2, Lon2.

dLat = D cos C in miles
dLon = D sin C / cos Lmid

where: Lmid = (Lat1 + Lat2)/2

Lat2 = Lat1 + dLat
Lon2 = Lon1 - dLon

Step-by-step example: Lat1 = 42° 00' N, Lon1 = 47° 00'W. Ship traveling 22 knots for 18h and 57m on course 266° true. Find Lat2, Lon2.

First we find distance travelled, D = 22 x (18h 57m) = 22 x 18.95 = 416.9 miles.

Then dLat = 416.9 cos 266° = 416.9 x -0.0698 = -29.1 miles = -29.1' [negative indicating southward movement]

Then we find Lat2 = Lat1 + dLat = 42° 00' - 29.1' = 41° 60' - 29.1 = 41° 30.9' N.

Next we find Lmid = (Lat1 + Lat2)/2 = (42° 00' + 41° 30.9')/2 = (42.00° + 41.52°)/2 = 83.52°/2 = 41.76°

Now we can find dLon in minutes-of-arc = D sin C / cos Lmid = 416.9 sin 266° / cos 41.76° = - 416.9 x 0.9976 / 0.7459= -557.5' [negative here means westward]

Now we can get Lon2 = Lon1 - dLon = 47° 00' - (-557.5') = 47° + 557.5' = 47° + 9° 17.5' = 56° 17.5' W.

Thus we end up at 41° 30.9' N, 56° 17.5' W.
That's very helpful Sam. Thanks for your explanatioN!
 
Mar 22, 2003
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This was his "Impression".
Yes, a very interesting statement when asked if he thought that Smith intended to go past the corner and then come back to the charted course line. He discovered the correct compass deviation error after he worked the 7:30pm star fix.

But we should consider the following. The ship's course was set to be steered by reference to the standard compass. There was no gyro compass available at the time to steer true courses. Over the period of setting the course to steer by and the next course to steer by, the magnetic variation heading westward would decrease as Jim rightfully pointed out. From turning near the corner point at 5:50pm 14 April to LAN 15 April the ship would travel some 417 miles, and the compass variation would have changed about 3° based on the International Geomagnetic Reference Model for 1912. So if they kept to a steady course by standard from the time they turned the corner to noon the next day, the true course followed would have shifted northward by almost 3° by time noon was reached the following day. I don't know what the practice was in 1912, but in aviation flight planning, if there was a significant change in variation between points A and B, we would use the magnetic variation at the median meridian rounded to the nearest whole degree. This may even explain why the 1st CQD was located on a true bearing of 264° from the corner if the intent was to make a true course of 265° to reach the noon location for April 15 (which would have been close to 56° 16'W). The median meridian would have been about 51° 38'W, and the variation there was about 1° less than what it was at the corner.

By the way, the Smith CQD, at 41° 44'N, 50° 24'W, is located 153 miles on a course of 264°T from the corner at 42°N, 47°W. If Smith took 11:45pm as the time of collision, and assumed that the vessel was making the same speed made good as she did the day before, 22.1 knots, then it would take 6h 55m to travel that distance from the corner. And here is where I believe a simple mistake of 1 hour took place, because the time difference from 5:50pm to 11:45pm is 5h 55m, not 6h 55m. A simple mental error in arithmetic made in the haste to work out a distress position. I just don't buy into the theory that the 8pm DR was 20miles too far westward. If we back out a one hour error from the Smith CQD we arrive at 41° 46'N, 49° 55'W, not far from the wreck site.
 
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Jim Currie

Member
Apr 16, 2008
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Funchal. Madeira
Jim, that was Pitman not Boxhall. I agree that Pitman's 5pm at the corner was ludicrous, but he got that by reversing what he knew by then. He took for granted that the ship reached Boxhall's CQD location at 11:46pm. The CQD is 145 miles from the corner. At Pitman's 21.5 knots it would take 6h 45m to get back to the corner. 11:46-6:45 = 5:01pm.
Simple reasoning, but wrong.
Yes. I pasted that from the transcripts. I meany to write "The Pitman states:"

Pitman stated:

"15174. I considered we went at least 10 miles further south than was necessary.15175. Do I understand you rightly that in marking the course at noon, the course was marked 10 miles further south than you considered necessary? A: - No. We had a certain distance to run to a corner, from noon to certain time, and we did not alter the course so early as I anticipated. Therefore we must have gone much further south.
15176. When did you alter the course? A: - 5.50.

He too was waffling. He must have known where the ship was before 8 pm because he also stated:

"Senator SMITH. : After making these observations, what did you do? It was then about 20 minutes to 8.
Mr. PITMAN: After that I started working out the observations....I was there alone until 8 o'clock....I did not finish them. Mr. Boxhall took over then and finished them."

He must have obtained a fix before 8 pm because he also said the ship was right on the track. The only way he would be able to determine this would be by calculating the course and distance back to the corner from a calculated fix position, i.e. if the course back to The Corner from the fix was North 65 East then he was right on the track. It would not be plotted.