Samuel Halpern
Member
I Agree. But the issue this year was the departure date, the overall cost, and the convenience of the port of departure for me. Maybe some other time I'll make it up there.
L
L. Marmaduke Collins
Member
Senator PERKINS.
When you were relieved on watch, Capt. Lightoller -
Mr. LIGHTOLLER.
I am not "captain."
Senator PERKINS.
You have a certificate as captain, have you not?
Mr. LIGHTOLLER.
Yes.
Senator PERKINS.
Then you are entitled to the honor.
Mr. LIGHTOLLER.
No; I do not claim the honor of the title "captain." I am plain "mister," as yet.
Even today, many merchant marine officers who hold a Master Certificate of Competency but have never had command of a ship, who sometimes work ashore as steamship inspectors, navigation instructors, etc., refer to themselves as ‘captain’.
I have had command of a ship, and I have had extensive experience in ship handling and navigating Arctic pack-ice. It is my opinion that the conclusions reached in the RMS Titanic: reappraisal of evidence relation to SS Californian have no merit. It is reminiscence of the testimony of Captain John Knapp, a United States Navy hydrographer, to the US Senate Titanic inquiry and the input of the nautical assessors to the British enquiry.
Excerpts:
1. INTRODUCTION
On 14 April 1912 at about 2340 hrs ship’s time the White Star liner TITANIC on her maiden voyage from Southampton towards New York struck an iceberg and was severely damaged.
The conclusions reached were that the Titanic was in the approximate position 41 47N, 49 55W when she struck the iceberg at 2345 hrs 14 April and that the Inspector considered that the Titanic was seen by the Californian, by her Masters and others. I think it is possible that she was seen, since abnormal refraction permitted sight beyond the extent that the horizon ordinarily would have been visible, but I think it is even more likely that she was not seen.
L
L. Marmaduke Collins
Member
Bill West
osted on Thursday, March 1, 2007 - 8:26 am:
You are incorrect on all 3 points!
1) If by a flattened sphere earth you are referring to the formula using Mean Lat and converting D. Long. into Dep. by the parallel sailing formula.
Mean lat. is the arithmetical mean between two latitudes and assumes the earth to be a plane surface.which gives approximate results only.
Those 1623/24 mile rhumb line routes are mid-latitude calculations for a spherical earth, they would be consistent with 1912 practice
2)
3) Middle latitude is the geographical mean between two latitudes and takes into account the convergency of the meridians on the surface of the earth. The middle latitude is the latitude in which the true departure lies. (Dep. = D.Long cos Mid. Lat.) It is found by applying to the mean latitude the middle latitude correction, which would be consistent with 1912 practice and testimony.
Collins
osted on Thursday, March 1, 2007 - 8:26 am: Bill
You are incorrect on all 3 points!
1) If by a flattened sphere earth you are referring to the formula using Mean Lat and converting D. Long. into Dep. by the parallel sailing formula.
Mean lat. is the arithmetical mean between two latitudes and assumes the earth to be a plane surface.which gives approximate results only.
Those 1623/24 mile rhumb line routes are mid-latitude calculations for a spherical earth, they would be consistent with 1912 practice
2)
Sam used round sphere calculations, mid-latitude calculations ??
3) Middle latitude is the geographical mean between two latitudes and takes into account the convergency of the meridians on the surface of the earth. The middle latitude is the latitude in which the true departure lies. (Dep. = D.Long cos Mid. Lat.) It is found by applying to the mean latitude the middle latitude correction, which would be consistent with 1912 practice and testimony.
Collins
B
Bill West
Member
LMC, your point numbers:
1 -For your query of the context of my point, the answer is that mid lat is not what I was referring to.
2 -You’re asking a confirmation of part of my point, the answer is yes.
3 -I didn’t make or challenge such a point.
Bill
1 -For your query of the context of my point, the answer is that mid lat is not what I was referring to.
2 -You’re asking a confirmation of part of my point, the answer is yes.
3 -I didn’t make or challenge such a point.
Bill
L
L. Marmaduke Collins
Member
Bill
Sam’s calculations give 1620 nm :
Collins
Were you referring to the surface of the earth as an oblate spheroid? If so, how would the calculations be consistent with 1912 [marine] practice and testimony?
1623/24 mile rhumb line routes are mid-latitude calculations for a spherical earth, which would be consistent with 1912 practice.
Sam’s calculations give 1620 nm :
Both calculations cannot be by mid-latitude method formula.
You wrote: Sam used round sphere calculations, they would be consistent with 1912 practice and testimony.
Collins
L
L. Marmaduke Collins
Member
Samuel Halpern: Posted on Thursday, May 31, 2007 - 2:32 am:
There is no significant difference between mid-latitude for spherical earth and mercator sailing calculations.
Example: Olympic Maiden Voyage: Daunt Rk to Corner 42/47
1st day
428.445 (428) nm Mercator sailing
428.572 Mid-lat formula
2nd day
535.652 (536) Mercator
535.675 Mid Lat formula
3rd day
542.680 (543 ) Mercator
542.719 Mid-lat formula
to corner:
173.616 (174 )Mercator
173.62 Mid-lat formula
TOTAL:
1680.393 Mercator
1680.586 Mid-Latitude
There is no significant difference between mid-latitude for spherical earth and mercator sailing calculations.
Example: Olympic Maiden Voyage: Daunt Rk to Corner 42/47
1st day
428.445 (428) nm Mercator sailing
428.572 Mid-lat formula
2nd day
535.652 (536) Mercator
535.675 Mid Lat formula
3rd day
542.680 (543 ) Mercator
542.719 Mid-lat formula
to corner:
173.616 (174 )Mercator
173.62 Mid-lat formula
TOTAL:
1680.393 Mercator
1680.586 Mid-Latitude
N
N. James Wright
Member
The difference I calculate to be the length of the ship plus 290ft. That does seem "insignificant".
Samuel Halpern
Member
Well there's that old proverb, "The devil is in the details." So lets show them.
Since Capt. Collins brought up the Olympic maiden voyage distances, lets take the reported run for the second day as an example since we have the starting and stopping coordinates from her log to work with, and therefore, we don't have to assume anything.
Starting coordinates were 50° 22'N, 19° 17'W. Ending coordinates were 47° 51'N, 32° 20'W.
Using Meridian parts we get a rhumb line distance of 535.652 miles (which rounds off to 536 miles) as Capt. Collins showed. (This result can easily be obtained from several on-line java scripts for those interested in doing so.)
Now let's use the mid latitude method to find the rhumb line distance.
What I did was 1st convert all degrees to minutes of arc.
Starting Lat coordinate is 50*60+22 = 3022' N
Ending Lat coordinate is 47*60+51 = 2871' N
Starting Lon coordinate is 19*60+17 = 1157' W
Ending Lon coordinate is 32*60+20 = 1940' W
The difference in latitude, dlat = 3022-2871 = 151' which is also the distance in nautical miles from north to south that the ship ran.
The middle latitude is the mean of the starting and ending latitude, or (3022+2871)/2=5893/2=2946.5' which expressed in degrees becomes 2946.5/60=49.11°
To get what is called the departure (east-west) distance in nautical miles, we first take the difference in longitude, dlon = 1940-1157 = 783'
and then multiply by the cosine of the mean latitude. This gives us 783*cos (49.11°) = 512.58 nautical miles in westward steaming.
We therefore have a southward distance 151 nautical miles, and a westward distance of 512.58 nautical miles using this method.
The total distance run is then found by taking the square root of the sum of the squares, or sqrt ( 151^2 + 512.58^2 ) = 534.36 nautical miles.
The difference in milage between the two methods is 535.62-534.36 = 1.29 miles.
By the way, the log card of the Olympic listed the run for the 2nd day as 534 miles. I got 534.36 which rounds off to 534. Meridian parts gives 535.65, which rounds off to 536.
Using this mid latitude method as described above, I get a distance for the 3rd day's run of 541.54 miles (which rounds off to 542 miles). The log card listed 542 miles. Meridian parts gives 542.681, which round off to 543.
Finally I got a distance from that noon to the corner of 173.22 miles, while meridian parts gives 173.616 miles.
So can we finally put this all to bed?
Since Capt. Collins brought up the Olympic maiden voyage distances, lets take the reported run for the second day as an example since we have the starting and stopping coordinates from her log to work with, and therefore, we don't have to assume anything.
Starting coordinates were 50° 22'N, 19° 17'W. Ending coordinates were 47° 51'N, 32° 20'W.
Using Meridian parts we get a rhumb line distance of 535.652 miles (which rounds off to 536 miles) as Capt. Collins showed. (This result can easily be obtained from several on-line java scripts for those interested in doing so.)
Now let's use the mid latitude method to find the rhumb line distance.
What I did was 1st convert all degrees to minutes of arc.
Starting Lat coordinate is 50*60+22 = 3022' N
Ending Lat coordinate is 47*60+51 = 2871' N
Starting Lon coordinate is 19*60+17 = 1157' W
Ending Lon coordinate is 32*60+20 = 1940' W
The difference in latitude, dlat = 3022-2871 = 151' which is also the distance in nautical miles from north to south that the ship ran.
The middle latitude is the mean of the starting and ending latitude, or (3022+2871)/2=5893/2=2946.5' which expressed in degrees becomes 2946.5/60=49.11°
To get what is called the departure (east-west) distance in nautical miles, we first take the difference in longitude, dlon = 1940-1157 = 783'
and then multiply by the cosine of the mean latitude. This gives us 783*cos (49.11°) = 512.58 nautical miles in westward steaming.
We therefore have a southward distance 151 nautical miles, and a westward distance of 512.58 nautical miles using this method.
The total distance run is then found by taking the square root of the sum of the squares, or sqrt ( 151^2 + 512.58^2 ) = 534.36 nautical miles.
The difference in milage between the two methods is 535.62-534.36 = 1.29 miles.
By the way, the log card of the Olympic listed the run for the 2nd day as 534 miles. I got 534.36 which rounds off to 534. Meridian parts gives 535.65, which rounds off to 536.
Using this mid latitude method as described above, I get a distance for the 3rd day's run of 541.54 miles (which rounds off to 542 miles). The log card listed 542 miles. Meridian parts gives 542.681, which round off to 543.
Finally I got a distance from that noon to the corner of 173.22 miles, while meridian parts gives 173.616 miles.
So can we finally put this all to bed?
L
L. Marmaduke Collins
Member
The middle latitude is the mean latitude 49° 6.5' minus the mid-lat correction 8'= 48°58.5'
Dlong 783 cosine mid- latitude 48° 58.5' = true departure 513.95 nautical miles, divided by
Dlat 151=tan course 73° 37.6'.
Dlat 151 secant course 73° 37.6' =535.675 miles. This mid-latitude calculation method for a spherical earth would be consistent with 1912 practice.
The difference in milage between the two methods is 535.675 -535.652 = 0.023 miles.(insignificant)
Samuel Halpern
Member
The difference from what I did and what Capt. Collins did in the mid lat calculations is that he used a "mid-lat correction" factor to get an adjusted number for the departure distance. He claims that this would be consistent with 1912 practice, and that is where I have to differ.
Let's look at the data.
For the run posted for June 17:
Log showed 534, I got 534.36 (534), Capt. Collins gets 536.675 (537)
For the run posted for June 18:
Log showed 542, I got 541.54 (542), Capt. Collins gets 542.719 (543).
Another example, take the rhumb line from noon June 19 (41-33N, 54-47W) to noon June 20 (40-41N, 66-50W):
Log showed 548, I get 547.2 (547), by meridian parts you would get 549.260 (549).
Take Voyage #2 westbound on rhumb line from noon Jul 17 (41-15N, 57-26W) to noon Jul 18 (40-35N, 68-49W):
Log showed 518, I get 517.7 (518), by meridian parts you would get 519.671 (520).
So is is clear, at least to me, that the method used by those officers working out the distances on the Olympic were not worried about such things as mid-lat correction factors in computing the middle lat. This is not even mentioned in Bowditch. Is is likely that they used transverse tables to get their results which require some interpolation of the data to be performed. I doubt they would have bothered to use a mid-lat correction factor even if they had those available.
In any event, the values that I obtained using the mid latitude method as I described in detail in the above post of Thursday, July 5, 2007 - 5:26 pm, produce results which agree quite nicely with the distances recorded in the log. To me, that is what counts.
L
L. Marmaduke Collins
Member
You wrote:
I did not say the correct mid-latitude formula -using middle latitude instead of mean latitude - was used by Olympic officers.
I said, and proved by calculations, There is no significant difference between mid-latitude for spherical earth and mercator sailing calculations.
This mid-latitude calculation method for a spherical earth would be consistent with 1912 practice, and was at the time of my D.Tp. examination for Master certificate in 1958.
Substituting mean latitude for middle latitude certainly does not give accurate results.
By not doing accurate calculations you incorrectly put Titanic 126 miles from the "corner" on Noon 14th.
Samuel Halpern
Member
I would rather say it does not give precise results.
A few definitions from Bowditch for the uninitiated reader if I may:
mean latitude. Half the arithmetical sum of the latitudes of two places on the same side of the equator. Mean latitude is labeled N or S to indicate whether it is north or south of the equator. The expression is occasionally used with reference to two places on opposite sides of the equator, but this usage is misleading as it lacks the significance usually associated with the expression. When the places are on opposite sides of the equator, two mean latitudes are generally used, the mean of each latitude north and south of the equator. The mean latitude is usually used in middle-latitude sailing for want of a practicable means of determining the middle latitude. [my emphasis] See also MIDDLE LATITUDE, MIDDLE-LATITUDE SAILING.
middle latitude. The latitude at which the arc length of the parallel separating the meridians passing through two specific points is exactly equal to the departure in proceeding from one point to the other by middle-latitude sailing. Also called MID-LATITUDE. See also MEAN LATITUDE, MIDDLE-LATITUDE SAILING.
middle-latitude sailing. A method that combines plane sailing and parallel sailing. Plane sailing is used to find difference of latitude and departure when course and distance are known, or vice versa. Parallel sailing is used to inter-convert departure and difference of
longitude. The mean latitude is normally used for want of a practicable means of determining the middle latitude, the latitude at which the arc length of the parallel separating the meridians passing through two specific points is exactly equal to the departure in proceeding from one point to the other. [my emphasis] See also MEAN LATITUDE.
It seems the officers on the Olympic used the practicable means to calculate the distances. The values that I obtained using the mean latitude for the middle latitude produced results which agree very well with the distances recorded in the log of the Olympic. To me, that is what counts, and why I feel quite comfortable in using it.
L
L. Marmaduke Collins
Member
You neglected to state the published date of your Bowditch. While Bowditch 1958 states :
The mean latitude is usually used in middle-latitude sailing for want of a practicable means of determining the middle latitude. It also states :If highly accurate results are required, a different method should be used.
Bearing in mind Titanic was a British registered ship with British navigating officers, and the year was 1912. Nicholl’s Concise Guide 1962 states :
The mean lat. Is the arithmetical mean between two latitudes and assumes the earth to be a plane surface; it gives results which are sensibly correct for short distances only
The middle latitude, however, is the geographical mean between two latitudes and takes into account the convergency of the meridians on the surface of the earth. The middle latitude is the latitude in which the true departure lies. It is found by applying to the mean latitude a correction given in the Nautical Tables.
Samuel Halpern
Member
You are arguing about which method is more precise here. I see no reason for this to continue. I'll say it again:
The values that I obtained using the mean latitude (without correction) produced results which agree with the distances recorded in the log of the Olympic.
The values that I obtained using the mean latitude (without correction) produced results which agree with the distances recorded in the log of the Olympic.
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N. James Wright
Member
If the method used by Sam, matches the distances recorded in the log - does that not show that Sam is using the methodology they used.
