exam question

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Big Pete
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Re: exam question

Postby Big Pete » Sat Nov 19, 2016 2:37 am

Thinking about the slip question, the time and speed are not necessary to calculate the slip, nor is the propeller diameter.#

The initial and final counter readings enable us to calculate the no. of revolutions the prop has turned, multiplying that by the pitch of the propeller gives us the theoretical distance the ship would advance, in whatever units the pitch is measured in.

In this case, as D. Winsor said Revolutions turned 731929-616729=115200

Propeller pitch is 20 ft, therefore theoretical distance advanced is 115200 x 20 =2304000 ft. Divide by 3 = 768000 yards.

There are 1,760 yards in a Statute Mile and 2,020 yards in a Nautical Mile, (UK, Imperial measurement) the question does not specify what type of Miles were "observed" so lets assume that as we are talking about ships, and it will therefore be Nautical Miles

Dividing our 768000 yards by 2020 will give us 380.19801 N.M. (I think the The 6080 in the Sothern's formula is wrong, as it should represent the no of feet in a Nautical Mile which is 2020 yards x 3 = 6060 ft)

Our observed distance was 404.16 presumed to be NM.

The difference is 380.198 - 404.16 =- 23.96199 this is the "absolute" slip in NM As the observed distance is, unusually, greater than the theoretical distance we have -negative slip as previously explained.

The % Slip is the Slip divided by the theoretical advance x 100.
-23.96199 /380.19801 x 100 = -6.3029%

If we assumed that the "Observed Miles" were Statute Miles, converting to Nautical Miles would give an observed distance of 404.16 x 1760/2020 = 352.139 NM

The Slip would then be 380.19801 - 352.139 = +28.059
The % slip would be + 28.059 /380.19801 x 100 = +7.38

Hope this helps someone BP.
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Big Pete
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Re: exam question

Postby Big Pete » Mon Nov 21, 2016 1:48 am

My apologies, I have just double checked the value of the Nautical Mile, and it looks as if we are all wrong! I found this: -

From Wikipedia, the free encyclopedia
Nautical mile
Unit system SI derived unit
Unit of Length
Symbol M, NM, or nmi 
Unit conversions
1 M, NM, or nmi in ... ... is equal to ...
meter 1852[1]
foot ≈6076.12
statute mile ≈1.15078
cable 10

Historical definition – 1 nautical mile

Graphic scale from a Mercator projection world map, showing the change with latitude

Visual comparison of a kilometer, statute mile, and nautical mile
A nautical mile is a unit of measurement defined as exactly 1852 meters (about 6,076.1 feet or 1.1508 statute miles). Historically, it was defined as one sixtieth of the distance between two parallels of latitude separated by one degree. Today it is an SI derived unit, being rounded to an even number of meters[2] and remains in use for both air and marine navigation[3] and for the definition of territorial waters.[4]

The derived unit of speed is the knot, defined as one nautical mile per hour. The geographical mile is the length of one minute of longitude along the Equator, about 1,855.325 m on the WGS 84 ellipsoid.

Contents [hide]
1 Unit symbol
2 History
3 See also
4 References
Unit symbol[edit]
There is no internationally agreed symbol.[1]

M is used as the abbreviation for the nautical mile by the International Hydrographic Organization[5] and by the International Bureau of Weights and Measures[1]
NM is used by the International Civil Aviation Organization.[6]
nm (the SI symbol for the nanometer) is used by the U.S. National Oceanic and Atmospheric Administration.[7]
nmi is used by the Institute of Electrical and Electronics Engineers[8] and the United States Government Publishing Office.[9]
History[edit]
The word mile is from the Latin word for a thousand paces: mīlia. Navigation at sea was done by eye[10] until around 1500 when navigational instruments were developed and cartographers began using a coordinate system with parallels of latitude and meridians of longitude. In 1617 the Dutch scientist Snell assessed the circumference at 24,630 Roman miles (24,024 statute miles). Around that time British mathematician Edmund Gunter improved navigational tools including a new quadrant to determine latitude at sea. He reasoned that the lines of latitude could be used as the basis for a unit of measurement for distance and proposed the nautical mile as one minute or one-sixtieth (
1
/
60
) of one degree of latitude. As one degree is
1
/
360
of a circle, one minute of arc is
1
/
21600
of a circle (or, in radians,
π
/
10800
). These sexagesimal (base 60) units originated in Babylonian astronomy. Gunter used Snell's circumference to define a nautical mile as 6,080 feet, the length of one minute of arc at 48 degrees latitude. Since the earth is not a perfect sphere but is an oblate spheroid with slightly flattened poles, a minute of latitude is not constant, but about 1,861 meters at the poles and 1,843 meters at the Equator,[1] with a mean value of 1,852.3 metres (6,077 ft). Other countries measure the minute of arc at 45 degrees latitude, giving the nautical mile a length of 6076 ft (approximately 1852 m).[10]

In 1929, the international nautical mile was defined by the First International Extraordinary Hydrographic Conference in Monaco as 1,852 meters.[1]

Imperial units and United States customary units used a definition of the nautical mile based on the Clarke (1866) Spheroid. The United States nautical mile was defined as 6,080.20 feet (1,853.24 m) based in the Mendenhall Order foot of 1893. It was abandoned in favour of the international nautical mile in 1954.[11]

The Imperial nautical mile, often called an Admiralty mile, or more correctly, an Admiralty measured mile, was defined by its relation to the Admiralty knot, 6,080 imperial feet per hour, so 1 imperial nautical mile is about 1,853.181 meters.[12] It was abandoned in 1970 and, legally, references to the obsolete unit are now converted to 1,853 meters.[13]

See also[edit]
It is always better to ask a stupid question than to do a stupid thing.

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D Winsor
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Re: exam question

Postby D Winsor » Mon Nov 21, 2016 5:42 am

In defense of how I answered the "Slip" Question
I assumed that where there were 4 possible answers given with the question A. 1.04% B. 1.29% C. -6.65% D. -11.04% with 1 of the 4 being correct, the candidate was supposed calculate and identify the correct answer given the information provided.
I also assumed that where I believed that the question was from a very old book or syllabus (probably written and based on American Standards) and where this and the previous question in the post used either Imperial and US Measurement systems. That the "Admiralty" or US measurement of a Nautical Mile would have applied.
Finally even though I didn't use the more modern means of calculating Propeller Slip I did come up with a result close to one of the answers provided. The result quite frankly surprised me but is possible if the ship was sailing down a long river like the St. Lawrence or the Mississippi with a 2 to 3 knot current pushing the vessel on a voyage of almost 24 hours between 2 ports was possible. I've also seen seen slip results of 75% or higher on some ships I've been on in the St. Lawrence when fighting both the current and tide going back up the river.
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Big Pete
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Re: exam question

Postby Big Pete » Wed Nov 23, 2016 2:13 am

D.W., No criticism was intended, Indeed I had to own up to making a mistake on the length of the Nautical Mile, I have had it in my head for years that the Nautical Mile was 2020 yards, and obviously I was wrong!! My point was that if you understand that slip is the difference between the theoretical advance of the screw and the actual distance travelled then, in this case you can use the total distances for the voyage more easily than reducing them to an average per hour.
Could be that the question was pre calculator days and the answers were all calculated on a "Guessing Stick" and none of them is to calculator accuracy, just 2 figure accuracy.
BP
.
It is always better to ask a stupid question than to do a stupid thing.

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Merlyn
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Re: exam question

Postby Merlyn » Wed Nov 23, 2016 4:57 am

Just passed a couple of these on a hundred foot super yacht in Gibraltar and couldn't resist wonderering how they would fare in the prop slip/ cavitation calcs?
See how far from the stern they are projecting about two feet.
You can stand on them to climb aboard.
No rudders just an adjustable shafting on a shock absorber type arrangement.
Pair of MAN's thro. ZF boxes gives about forty knots.
Six bladers surface props, get the leading edges ground flat and thereby losing half the blade surface contact area with the water.
To obtain maximum efficiency they operate best when half in / half out the water.
Pig to manouvre when going astern plus the stern to method of mooring in the Med. leaves the blades way open to chewing lumps out of the quay wall.
Weird set up but maybe the calcs. have to be the same?
D.W?
IMG_0580.JPG
Blading
IMG_0581.JPG
Six half blades?
IMG_0582.JPG
Shafts are linked together
IMG_0583.JPG
Shafts go up and down
Remembering The Good Old days, when Chiefs stood watches and all Torque settings were F.T.

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D Winsor
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Re: exam question

Postby D Winsor » Thu Nov 24, 2016 5:13 am

Big Pete wrote:D.W., No criticism was intended, Indeed I had to own up to making a mistake on the length of the Nautical Mile, I have had it in my head for years that the Nautical Mile was 2020 yards, and obviously I was wrong!! My point was that if you understand that slip is the difference between the theoretical advance of the screw and the actual distance travelled then, in this case you can use the total distances for the voyage more easily than reducing them to an average per hour.
Could be that the question was pre calculator days and the answers were all calculated on a "Guessing Stick" and none of them is to calculator accuracy, just 2 figure accuracy.
BP
.

BP You are right the question was probably designed to be answered with as you call a "Guessing Stick" or Slide Rule. It has been 43 years since I first learned to do "Slip" calculations and I did use at the time, plus I still have and know how to use, my slide rule. At that time the Electronic Calculator you buy for a dollar today cost up to $1000, making them well out of reach for the average "Poor" Cadet, plus at that time Transport Canada still did not permit the use of any kind of electronic devices such as calculators when doing exams. Today I show the slide rule to family members studying engineering and they are amazed at its functionality. I also suggest the next time they happen to watch the Movie Apollo 13 with Tom Hanks to take note of the file footage of mission control as the engineers are using slide rules not computers or calculators to do their calculations.
Last edited by D Winsor on Fri Nov 25, 2016 3:55 am, edited 1 time in total.
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JK
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Re: exam question

Postby JK » Thu Nov 24, 2016 6:34 am

Merlyn, for the life of me I can't understand why your pictures post upside down...I know you aren't in Australia!

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Merlyn
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Re: exam question

Postby Merlyn » Fri Nov 25, 2016 3:56 am

Perhaps I charged the battery up wrong way round?
You know, reversed polarity?
Remembering The Good Old days, when Chiefs stood watches and all Torque settings were F.T.

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JK
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Re: exam question

Postby JK » Fri Nov 25, 2016 4:51 am

They show right side up on the IPad but not on the desktop. It makes for some puzzlement at what exactly I'm looking at. LOL

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Merlyn
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Re: exam question

Postby Merlyn » Fri Nov 25, 2016 5:17 am

Weird, not very good photos though, could be a macerator pump in the offing?
Research seems to indicate the same formulae applies here re slippage/ cavitation calcs ?
Partime props, best efficiency half in / half out penetrations.
Best you study these as when you retire and draw out you might end up with one of these?
Stand on props I call them.
Remembering The Good Old days, when Chiefs stood watches and all Torque settings were F.T.


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