You can submit here your questions or problem that you wanted to be solved. Then I will try to answer it ,or much better if the readers will try to share their solutions.You can use the comment box to write a particular question.
And if your questions are answered here, please return a compliment by hitting the "like" buttons and/or share this blog on your Facebook.
Thank you and welcome.
And if your questions are answered here, please return a compliment by hitting the "like" buttons and/or share this blog on your Facebook.
Thank you and welcome.
Here ls a question from one of the readers on this blog.
ReplyDeleteIf your mercurial barometer reads 1033 mb and the temp is 13 deg. C, what is the correct reading at 55 deg. North 150 deg. West.
Anyone who knows how to solve this problem?
This comment has been removed by the author.
Delete2.You desire to make good a true course of 129°. The variation is 7°E, deviation is 4°E, and gyrocompass error is 2°W. An easterly wind produces a 4° leeway. What is the course to steer per standard compass to make good the true course?
DeleteT/C 129 minus 2 W Gyro error = 127
DeleteT/C 127 minus 4 easterly wind leeway = 123
T/C = 123
Var = 7 E (-)
Mag.= 116
Dev.= 4 E (-)
PSC = 112 (Final Answer)
Your vessel departure on August 10, 2021 time=1820 hrs at position to Lat=49deg -40mins North, Long=004 deg-15 mins West. And Arrive to Lat =48 deg-15 mins North, Long=006 deg-30 mins West. Speed=16 knots.
DeleteYOUR VESSEL DEPART ON APRIL 12, 2022, TIME=1330HRS AT SPEED=16 KNOTS. VESSEL DEPARTURE POSITION TO WAYPOINT 0=LAT=50 DEG-17 MINS. N, LONG=006 DEG, 18 MINS. W, POSITION TO WP 1= LAT=50 DEG-11 MINS. N, LONG=006 DEG-02 MINS. W. TO WP 1, YOU ALTER YOUR COURSE TO 179 DEG, DISTANCE TO WP 1 TO WP 2=24 NM.AND YOUR ARRIVAL POSITION TO WP 3 OF LAT=48 DEG-48 MINS N, LONG=004 DEG-45 MINS W.
DeleteYou have steamed 925 miles at 13.5 knots, and consumed 181 tons of fuel. If have 259 tons of usable remaining, how far can you steam at 16 knots.
DeleteDear Sir,
DeleteI am now having my review and I found it difficult to solve this particular problem:
A vessel displacement is 60110 tons has a TPC 63 TONS and MTC 888.7 kilotons. Draft fwd reads 10.0m and draft aft 10.2m. The trim table for 100 KT shows Fore 3.43 and aft -0.32. Calculate the new draft Fwd and aft if you are to load 300 tons into cargo hold no. 3?
A million thanks to whatever assistance you can extend.
A vessel displacing 18030 tonnes, Kg 7.6m Km 8.45m is upright.
DeleteA lift of 85tonnes Kg 13.2m, 6.0m to port of the centre line is to discharged into barge 12.6m to starboard of the vessel centre line by the ships heavy derrick the head of which is 27m above the keel.
Calculate (i) The maximum angle of heel
(ii) The final Gm after Discharging Weight
(ii) The final angle of heel after discharging
Referring to Light list, a light has a nominal range of 14 miles and 12.8 meters high. If the visibility is 6nm and your height of eye is 6.0m, at what approximate distance will you sight the light? why the answer is 10 nautical miles
DeleteHow about this one? Can anybody teach me how to solve this?
ReplyDeleteA ship "A" is on the equator steering 090°T at 16 knots ; while a ship"B" is on the parallel of north latitude, steering 270°T at 12 knots. When "A" makes Dlo of 1', "B" makes a Dlo of 48'. Calculate the latitude of "B".
Ans: Lat 20° 22'
Follow this link to see my answer.
Deletehttp://oic-nwreviewer.blogspot.com/2014/05/question-from-readers-2.html
How about This one A VESSEL IS ON A COURSE OF 079 DEGREE T AT A SPEED OF 18 KTS. AT 0218H, A LIGHTHOUSE IS OBSERVED BEARING 111 DEGREE T. ON WHAT 2ND BEARING MUST THE LIGHTHOUSE AGAIN BE OBSERVED SO THAT THE DISTANCE RU BETWEEN BEARING IS EqUAL TO THE DISTANCE OFF WHEN ABEAM OF THE LIGHTHOUSE?
DeleteWhile in position Lat. 16° 30.5’ N Long 116° 45.5’E at 2030H LT (GMT+8) on May 20, 2022 star POLARIS was observed at 000° from the port side repeater. Ship’s heading at the time was 225° at Standard 220° . Var is 6°E.
Delete
DeleteA vessel from Lat. 35*27’ N, Long. 139*39’E sail to Lat. 37*48’N, Long. 122*20’W. Find the GCD, D’lat., D’long.,Initial Course Angle, Initial Course True, Final Course Angle and Final Course True.
one end of a ten foot ladder is four feet from the base of a wall. how high on the wall does the top of the ladder touch
DeleteA 7000 ton displacement tank ship carries two slack tanks of alcohol with a SG of 0.8 each tank i 50 ft. long 30 ft. wide. What is the reduction in GM due to free surface with the vessel floating in sea water, SG is 1.026?
ReplyDeleteHello, I answer your question. You can see it by clicking the link below.
Deletehttp://oic-nwreviewer.blogspot.com/2014/05/question-from-readers-3.html
Thank you sir!
DeleteYou've been a big help to those of us who are doing self-review.
Kudos to you sir @Nauta!
It's nice to hear people thanking us for our effort to help.
DeleteYou're always welcome
A ship sails from Lat 14°16.2’S with a course of 048° (T) with a speed of 11.6 knots. What is her latitude after 45 minutes? Can you solve this one sir
DeleteIf a pump can empty a tank in 16 hrs, another pump can empty the same tank in 8 hrs, and another can empty this tank in 12 hrs. How long will it take to empty it if the 3 pumps are working together on this tank.
ReplyDeleteHello!
DeleteSee my answer on this link
http://oic-nwreviewer.blogspot.com/2014/05/question-from-readers-4.html
What is the TPC of a box shaped vessel 80m x 14m floating in SW at an even keel draft of 4m?
ReplyDeleteTPC = (Lenght x Breadth x Block coefficient x Density of SW) ÷ 100
Delete= (80 x 14 x 1 x 1.025) ÷ 100
= 1148 ÷ 100
TPC = 11.48 is the answer
Hello can i ask this one?
ReplyDelete360 tons of cargo are loaded in a vessel 150' forward of the tipping center; tons per inch are 50; breadth of the ship 60'; block coefficient .75, draft before loading 20'00" F and 22'06" A. Find the draft after loading.
The answer here is
22'04.5"F 21'03.5" A
I just want to know how to solve it.. Thanks!
Hello Nigel,
DeleteSorry for the late reply. I just read your question today because I no longer have a computer nor a tablet. So that means that I may not post anything new until I have a computer or tablet.
However, I copy your question in paper, and I will try to answer it. I will tell my answer next time I have an opportunity to be online.
Thanks for understanding.
This comment has been removed by the author.
DeleteEdit: Thanks! I completely understand. Best of Luck!
DeleteNigel, I find it hard to answer your problem. Or I should say I cannot answer because there is no MTI in the problem. Nor we can solve for the MTI.
DeleteA ship has LBP 215 m, TPC 59.05 T, MTC 767.41 T-m, LCB 7.08 m with displacement of 31979 T and floating in a dock water density of 1.022 with draft 4.18 m FWD, 7.69 m AFT. What will be her draft n sea water considering there is no change in displacemet.
ReplyDeleteHi,
DeleteAs I said to Nigel, I no longer have a means to post but I copied your question and I will try to answer it.
Hi forgive my late reply.
DeleteMy answer is 4.16 m forward draft, and 7.67 m draft aft. See my latest post for my solution on Question From Readers #6.
a vessel displacement is 55713 tons has tpc of 62.4 tons and mtc of 863.2 kilotons. draft forward reads 9.30 m and draft aft 9.50 m. the trim table for 100 kt shows fore 4.76 and aft (-1.72) calculate the new draft fore and aft if you are to load 150 tons into cargo hold no. 2
Deletewhat is the formula for permeability?
ReplyDeletePermeability= Broken Stowage Factor x 100% / Stowage Factor
DeleteThanks for this Nigel.
DeleteMaraming Salamat.
You are on a box-shaped vessel about to enter a river. At the mouth of the river where the relative density is 1.018 ton/m? your draft was 5.80 m. What would be the approximate keel clearance in the river dredged to a depth of 8.50 m if its density 1.005
ReplyDeleteAnswer is 2.63 m.
DeleteSee my new post to see my solution.
a box shape 24m x6m x 3m displaces 150 tons of water. Find the draft when vessel is floating in salt water..?
ReplyDelete
DeleteHello,
My answer is 1.02 meter.
I'm not sure of that, though. So will you please share to us the answer?
I will answer this question if I can barrow again a laptop. hehehe
ReplyDeleteKeep coming.
Good day sir,
ReplyDeleteCan you please help me solve this problem that I've encountered on my review?
Question:
What distance will a weight of 100 tons have to
be moved in order to remove a 3 degrees list from
a vessel? Displacement is 9500 tons and GM is 3.0 feet.
Thanks in advance! Salute to all of the guys behind this website!
Its 14.9 ft.
DeleteFormula :
Tan list = (weight x distance) / (displacement x GM)
Derive to find distance;
Distance = (displacement x GM x tan list) / weight
D= (9500 tons x 3 ft x tan 3) / 100 tons
Cancel tons so ft will remain.
D= 1493.62 / 100
Distance is 14.94
Thanks Nigel for your solution. You've been a help to others.
DeleteA ship is inclined by moving a weight of 30 tons a distance of 30 ft. from the centerline. A 28-foot pendulum shows a deflection of 12 inches. Displacement including weight moved is 4,000 tons. KM is 27.64 feet. What is the KG?
ReplyDeleteMy answer is 21.19 ft. See my solution on my post Question From Reader #8. Let me know also what is the answer according to your review material. OK? Thanks.
DeleteA vessel's draft is 14-11 fwd and 14-07 aft. She loads 1,470 tons which is 40 feet aft of the tipping center with a TPI of 60 and a MTI of 1335. What is her new draft?
ReplyDeleteSee my post Question From Readers #7.
DeleteYou have steamed 1134 miles at 10 knots, and consumed 121 tons of fuel. If you have to steam 1522 miles to complete the voyage, how many tons of fuel will be consumed while steaming 1951 miles?
ReplyDeleteOn the categories of problem, click "Engine Related Problems" to see similar questions of yours.
DeleteThis comment has been removed by the author.
ReplyDeleteSir to find true wind speed by the formula from your website doesn't work for other type of wind problem?????Please let know if I m wrong?Thank u
ReplyDeleteCourse 045degree speed 15 knots.Apparent wind 100degree at 20 knots.Find the direction and speed of true wind. ANSWERS in this case is 147degree at 17 knots.
The formula given in this blog is tested. However, there are problems in which no matter how you solved with the correct formula, you can not find the correct answer among the choices.
DeleteI encountered many of it during my review. I asked my instructor about it, and he said you can not do anything about it but to select or choose the answer as told by your review material even though it is not the correct answer. We called it "PRC" questions.
GOOD DAY SIR,
ReplyDeleteYour blog is so very helpful to us , and i am very thankful one of my friend told me about this blog and i have some questions to ask about navigation.
1. AT 0800 zone time, on 15 april, your vessel is heading west in position LAT. 15?10? N, LONG. 165?15? W at a speed of 22 knots. The distance to your destination at LAT. 15?10?N, LONG. 135?15? E is 3600 nautical miles. What is your ETA???
ANS: 2339, 22 APRIL
2. You are on a voyage from Limoy, Costa Rica , to Los Angeles USA. The distance pilot to pilo is 3150 miles. The speed of advance is 14 knots. YOu estimate 24 hours for the Panama CANAl Transit. If you take departure at 1836 hours (ZD+6) on 28 January, find your ETA (ZD +8) at LOs Angeles?
ANS: 1336, 08 Febuary
3. At 1820 LZT, on 21 MArch 2007, you depart San Francisco at LAT. 37? 48.5? N, LONG 122? 24? W (ZD+8). You are bound for Melbourne, LAT 37? 49.2? S, LONG 144? 56? E and you estimate your speed of advance at 21 knots. The distance is 6,970 miles. What is your estimated zone time of arrival at melbourne?
HOPE YOU WILL ANSWER THIS QUESTIONS SIR .
THANKZ.....!!!
Hi Angelo!
DeleteRegarding to your question # 1, my answer is 0338H 22 April.
I will tackle your remaining questions later these days if I can since I have no longer have a personal computer.
I feature your question as Question From Readers #9. Here's the link and see my solution there http://oic-nwreviewer.blogspot.com/2014/08/question-from-readers-9.html.
Thankz a lot sir ...
DeleteHello,
DeleteWill you please check again your reviewer? Is the answer really 1336 8 FEB?
With regard to your question number 2, my answer to that is 0136 8 February is your LZT at Los Angeles.
Anyway, I have to feature this problem, an let's hope some readers may solve this problem.
Delete
THanks Much sir!!!
DeleteAngelo,
DeleteYou're welcome. How's the review?
good day sir.. i would like to ask an example on how to solve regarding to this problem below.. any example related to this question..thank you sir and godbless.. "Which of the following bearings of two fixed and charted objects will give a good crossing angle between 2 LOPs if taken at nearly the same time?"
ReplyDeleteIt would be better if you would provide a specific problem because I have no means to access my database.
DeleteGood day sir!
ReplyDeletecan you please show me the solution of solving this problem:
A vessel 580 ft x 60 ft, with a waterplane coefficient of 0.84 is floating in freshwater at a draft of 21 feet. How many long tons shall it take to increase the mean draft by 1 inch?
Answer (According to our reviewer): 67.6 tons
Thanks a lot sir and God bless.
Sure bro,
DeleteLook at my post Question From Readers #11.
I-ampo lang kog apil bai nga madawat ko sa gi-aplayan nako nga shipping. Thanks
hi gud eve... can you pls help me with this? DATE 16 MAY 2007 00H31M35S GMT POS. LAT 9 28 N LONG 117 10 E. GYRO HEADING 37 SPEED 16KTS. MAGNETIC COMPASS 43 VARIATION 9W GYRO BEARING SUN 75.3 can you pls also show how you did it. many thnks
ReplyDeleteSeems to m incomplete. What is asked in the problem?
Deleteahh ok sorry. T/AZ. GYRO ERROR. DEVIATION. I would like to compare my ans to your ans. Many thanks in adv. coz i have nobody here to help me. God bless
DeleteThe deviation is 3° E
Deletegud eve... i need you help with this. SHIP AT 0800 OCT 10 2011 LAT 9 2 N LONG 116 49.5E TRUE COURSE 38 SPEED 16 KTS FIND DEADRECKONING POS. AT 1600 OCT 10 2011
ReplyDeleteI will solve this later if I have time. But for now I would just give a general concept.
ReplyDeleteThis is kind of a plane sailing. If you try to figure it out, it would constitute a plane triangle.
Calculate the distance traveled, and the result would be your hypotenuse. Use 38 degrees as your angle. With this given data, you can solve your new position.
Hi good day. my ans. to this problem is Lat. 10 49.2 N Long. 118 14.7 E. i dont know if this is correct. and also do you have a problem solving regarding meridional parts? and problems for requiring the speed esp. long distance voyage. exp from shanhai to balboa? thank you very much. im very thankful that i found your blog. t.c always bro..
ReplyDeleteSee my solution of your problem on this link
Deletehttp://oic-nwreviewer.blogspot.com/2014/09/question-from-readers-12.html
Good day, Calculate sag correction basis TPC=41.06, MTC=481,LBP=163.6 and estimated sag = 4cm. Ans. 117.2 MT
ReplyDeleteHopefully somebody has the answer.
DeleteFind the cargo to be loaded if initial draft aft is 10.873m, Trim Corr. = 14.408, TPC = 38.37 and final draft aft = 10.9m.
ReplyDeleteAns. 89.19tons
See my post Question From Readers #14
Deletea tank measure 3 feet x 32 feet x 40 feet. how long will it take to empty the tank if the pumps discharge rate is 120 gallons per minute... is the answer 59 mins?
ReplyDeleteIt's 3 hrs 59 min 21 sec to empty the tank. See my post "Question From Readers #15".
Deletehow many tons of boxes measuring 3 feet by 2.2 feet by 5 feet and weighing 560 pounds can be stowed in a space having a cubic capacity of 28500 cubic feet assuming a broken stowage of 15%? answer: 161.5 tons?
ReplyDeleteAs to this and the rest of your questions below, I have solved these type of problem. Just use the search box on this blog and type your query. Or you may click "Ship Stability" and "Cargo Operation" on the Categories of Problem to find similar problems.
Deletea vessel has a mean draft of 25'-00", her displacement is 10000 tons and the TPI is 42 . Find the increase of draft in a water density of 1010 kgs/m3 answer 3.5"
ReplyDeletethe displacement in saltwater is 23,900 tons and the tpc is 31.44, permitted freshwater sinkage is 19cm. what is the equivalent displacement in freshwater? answer: 23,894 tons
ReplyDeletea tank has a capacity of 90,000 barrels . you load 88,200 barrels at 60degF of a commodity having an API at 60degF of 58.3. to what tempreature this commodity bea heated before the tanks overflows? answer: 93 deg f
ReplyDeletethe fwa being 6" for vessel M/V seafarers. what is the increase of draft allowed in water oF RD 1.010?
ReplyDeleteSee Question From Readers 16
Deletehttp://oic-nwreviewer.blogspot.com/2014/09/question-from-readers-16.htm
Calculate the righting arm of a vessel whose GM is 0.65 m when inclined at an angle of 12??
ReplyDeleteSee Question From Readers #17
DeleteIf you've been helped, hit the like button in this blog.
A vessel with a beam of 72 feet, has an average rolling period of 18 seconds. What would be her metacentric height?
ReplyDeletemetacentric height is 3.09 feet.
DeleteSee Question From Readers #19
A vessel with a beam of 72 feet, has an average rolling period of 18 seconds. What would be her metacentric height?
ReplyDeleteit's been a really big help to me,. but i can't find the like button...
ReplyDeleteThere's a "Share It" button in the lower right side.
DeleteThis comment has been removed by the author.
ReplyDeleteA vessel's displacement is 2,400 tons and her KG is 10.8 meters. What is the new KG if a weight of 50 tons already on board is raised 12 meters vertically?
ReplyDeleteAdrian,
DeleteMy answer is 11.05 meters. See "Question From Readers #20
This comment has been removed by the author.
ReplyDeleteYour vessel is making way through the water at a speed of 10 knots. Your vessel traveled 30 nautical miles in 2 hours 30 minutes. What current are you experiencing?
ReplyDeleteans: following current of 2.0 knots
can i ask what is the solution of this problem sir?
Hello Kim,
DeleteI have featured already this kind of problem. This is the link http://oic-nwreviewer.blogspot.com/2014/03/what-current-are-experiencing.html
If the link doesn't work, open the blog archive March 2014 and click the post "What Current You Are Experiencing".
DeleteA cargo of oil has a coefficient of expansion of .0005 per degree F. If this cargo is loaded at 70?F, and a cargo temperature of 90?F is expected at the discharge port, how many barrels would you expect to unload if you loaded 10,000 barrels?
ReplyDeleteanswer: 10100
can you please show me the solution?
thanks in advance
See post "How Many Barrels Would You Expect To Unload" (1 Nov 2014)
DeleteWhat size of manila line is required to hold a weight of 932 lbs., if you use a safety factor of six?
ReplyDeleteYour Answer :: 3.0? Correct Answer :: 2.5?
what is the solution of this
From my solution, the answer is 2.49265. Yes the correct answer is 2.5, not 3.
DeleteSince it's late in the evening now, I'll show you my solution next time. I hope it's okay for you.
Kim, see my solution on "What Size Of Manila Rope Is Required" (08 NOV 2014).
Deletea rough guide, in the absence of an accurate meteorological data, if the corrected barometer reading is 5 hPa (0.15 in.) lower than the local normal and the wind force is about 8, the tropical cyclone's center will be about how far away? answer:100nm
ReplyDeletethanks in advance
Hello Mejanuss,
DeleteI never encountered this problem before. Hopefully someone who knows this will post their solution here.
I hope u can help me with this problem I encountered on my review material, " if the distance between 2 consecutive isobars( drawn at intervals of 4mb) is found to be 100nm and the latitude is 50 degrees, what is the Geostrophic wind speed?" thankyou in advance :)
ReplyDeleteth anchorage is loacated 20nm north of a vessel with a speed of 15knots,A current hava a set of 3knots E,calculate the course to steer in order to drop her anchor to the charted anchorage on the spot,ans:349degrees
ReplyDeleteYou are underway on course 50*T and speed of 12knots. the eye of the storm is moving towards 265*T at 22knots. if you maneucer at 12knots to avoid huriccane. what could be the cpa? ans 63miles
ReplyDeletecan i ask a question?
ReplyDeletewhat is the meaning of panofsky-brier ?
what is the exact formula ?
As to the meaning of Panofsky & Brier, you better read a standard reference online.
DeleteWith regard to the Panofsky & Brier formula, we have it here on this site. You can use the search box and then type panofsky and brier. Or you can scan through the blog archive.
Good day sir.can you help me sir for this question..
ReplyDeleteVsl displacement 10,000 tons.350 ft long and has a beam of 55 ft.You have timed it's rolling period to be 15.0 seconds.What is your vsl approximate GM?
Thank you in advance for the answer..
DeleteI solved that kind of problem in this blog. You can go through the arhcive and see my post on October 1, 2014. Here's the link http://oic-nwreviewer.blogspot.com/2014/10/question-from-readers-19.html
This comment has been removed by the author.
ReplyDeleteOn 26 July your 1030 ZT DR position is LAT 18*25'N, LONG 51*15'W. You are on course 231*T, speed 15 knots. Determine your 1200 position using the following observations of the sun.
ReplyDeleteAnswer: LAT 18*07.2'N, LONG 51*30.4'W
Can anyone please show the solution. Thank you
It is a plane sailing, I think. My answer is not the same as yours.
DeleteGood Day Sir,can you help me for this question.
ReplyDeleteA ship has to load copper ingots in pallets weighing 1.1MT each.The dimensions given are 0.8mx0.6mx0.9m.The bale capacity of the lower hold is 3200m3;it has a height of 8m.the permissible load density is 9mt/sq.m.Find how high can the pallets be loaded and also the number of pallets as well as the total load.Additionally how would you go about spreading the weight.
The No.3C cargo tank of a ship has a depth of 26.5m from the ullage port.Water cut(level check) was taken and it was found that the free water was 11cm.What would the ullage be if measured by an electronic ullage tape.
ReplyDeletecan i ask what is the solution.
Thank you
Good day sir ,can you help me to understand what is beaufort scale is 0.836 cube root of b is a formula ? Can you please me an example how thanks
ReplyDelete@0900 UTC, a life raft is reported to be in lat 30deg 56.4min S long 000 deg 25.6E. A rescue ship reports that it's ETA at the vicinity is 2130 UTC. The rescue ships navigator calculates that wind and ocean current will cause the life raft to drift 345 deg at 3knots. Calculate the expected position of the life raft when the rescue ship is due to arrive.
ReplyDeletehow many tons of boxes measuring 3'06" x 3'06" x 6'6" and weight 750 lbs can be stowed on lower deck having a cube capacity of 29300 cu. ft. using broken stowage of 15 ft. ? tnx
ReplyDeletea ship has to load copper ingots in pallets weighing 1.1 mt each. the dimensions given are 0.8m * 0.6m * 0.9m. the bale capacity of the lower hold is 3200m3; it has a hieght of 8m. the permissible load density is 9mt/sqm. find how high can the pallets be loaded and also the number of palletsas well as the total load. additionally how would you go about spreading the weight?
ReplyDeleteA vessel steams 720nm on course 058ºT from LAT 30º06.0'S, LONG 031º42.0'E. What are the latitude and longitude of the point of arrival by mid-lattude sailing?
ReplyDeleteat about 1300/26th feb 2014 vessel sailed from port Saranggani is along Lat/13-28N, Long/125-23E and bound for guam is along Lat/12-28N,Lomg/144-36E at a speed of 12.8 knots. That the vessel upon arrival at guam. Vessel schedule to bert upon her arrival then affter 10days exactly including discharging,vessel schedule to depart Guam with same Lat/Long above, heading toward Jayapura port,Papa New Guinea along Lat/02-305,Long/140-SE a.)Illustration showing DLAT,DLO,DMP,DEP AND DISTANCE b.)DLAT AND DLO between saranggani and guam c.)course and distance d.)steaming time between sarangani and guam d.)ETA at Guam e.)ETD guam using above 10 days stay at guam including discharging f.)DLAT and DLO between guam and Jayaputra port g.) course and distance h.)steaming time between Guam and Jayapotra,New Guinea i.)ETA Jayaputra New Guine using speed of knots
ReplyDeleteSir na Solve nyo na po ung problems na to thank you 😇
DeleteThe scale of a Mercator projection is 4 inches equals 1°LONG.
ReplyDeleteWhat is the expansion in inches between the 60th and 61st parallels?
NEED THIS ASAP
You wish to measure the distance on a Mercator chart between a point in latitude 42°30'N
ReplyDeleteand a point in latitude 40°30'N.
To measure 30 miles at a time you should set the points of the dividers at __________.
Can anyone please show me how to compute this one manually?
What are the common errors in a gyro compass?
ReplyDeleteAt about 1300/26th feb 2014 vessel sailed from port saranggani ls. Along Lat/05-24N, long /125- 23E and bound for guam is Along lat/13-28N, Long/144-36E at speed of 12.8 knotsm that the vessel upon arival at guam.
ReplyDeleteVessel sched to berth upon her arival then after 10 days exactly including discharging, vessel sched to depart Guam w/same Lat/Long of above, heading towards hayapura port, papua new guinea along Lat/02-305, long/140-05E.
A. Draw a triangle w/ illustration showing DLAT,DLO,DMP,DEP and Distance.
B. DLAT and DLO between sarangani ls and Guam ls.
C. Course and Distance
D. Steaming time between sarangani ls and guam ls.
E. ETA at Guam
F. ETD at Guam using above 10days stay at Guam including discharging.
G. DLAT and DLO between Guam and Jayaputra port.
H. course and distance.
I. Steeming time between Guam and Jayaputra port, new guinea.
J. ETA jayaputra port using a speed of knots.
Yan mga sasagutin.
Please Reply me thank you so muchhh
DeleteAt 0800H, two ships was observed to be 33 miles apart on opposite courses. One ship was running at 12 knots while the other was making 10 knots. At what time of the day will they be abeam from each other?
ReplyDeletePatulong please. Nagrereview po ako for the Pre-Qualifying exam
The answer to your question is 0930H. That is the time the ships will be abeam to each other.
DeleteThe vessel is floating at dock with 11.5m forward and 12.5m aft. Solve for underkeel clearance where the depth of the water at the area is 13m
ReplyDeletePatulong naman po please. Thank you
Hi in have a question regarding longitudinal stability of vessel.
ReplyDeleteHere's the question :
A box-shaped vessel 40m * 6m * 3m is floating in salt water on an even keel at 2m draft forward and aft. How do I find the new draft if a weight of 15 tonnes is discharged from a position 6m from forward end.(use BML for mctc calculation)
i try to use units in mtrs with the same values but can not arrive the same answer on rolling period as shown pls explain further
ReplyDeleteIm trying to get the longitude of arrival? Here is question- u depart lat 49'45N long 06'35W and steam 3599 miles on course 246.5 T. What is the longitude of arrival?
ReplyDelete1. Given a compass course of 045°, variation 7°E, Magnetic course 049° and the wind blows in SE direction. If the leeway is 5° find the Dev. True co. and illustrate the course to steer(CTS).
ReplyDelete2.If the current set on SE direction and your co. is 225°T, find the leeway if course to steer is 218°.
3. Based on problem # 2, if the current set to NW direction and the leeway increase to 9° Find the Course made good( CMG).
A ship has a Summer Load line mark at 5m 80cm. Her FWA is 140mm and with 21.82 tons as TPC. The loading port has RD 1.007 and her draft at present is 5m 74cm. How much cargo could she load to be on her Summer mark of 5m 80cm on reaching seawater, allowing for the 28 tons of fuel to load, assuming that the TPC is unchanged?
ReplyDeleteA vessel floating in salt water has the following particulars:
ReplyDeleteDisplacement 1800 tonne Length B.P. 220.00m
LCB 100.00m forward of AP
LCF 120.00m forward of AP
MCTC 200 TCP 23
Draughts: Forward 7.85m Aft 8.55m
The vessel has two bunker tanks, the forward tank has its centroid 205.00
forward of AP and the after tank has its centroid 75.00m forward of AP.
Calculate:
(a) The amount of fuel to transfer between the bunker tanks in order to arrive
alongside at a Fresh Water berth on an even keel; (b) The arrival draughts fore and aft.
A box ship vessel has departed from a fresh water dock density with a draft fwd 7.05 and draft aft 8.00.Determine the draft when the vessel is out to sea after passing the sea buoy.
ReplyDeleteWhat is the approximately velocity of wind at low pressure area,If the storm central pressure 950 (hPa) .
ReplyDeleteThis kind of question is already featured on blog. Click the link below. You can find the answer there for it is a similar question to yours.
DeleteHere is the link https://oic-nwreviewer.blogspot.com/2013/12/panofsky-brier-fomula-for-wind-speed.html?m=1
https://oic-nwreviewer.blogspot.com/2013/12/panofsky-brier-fomula-for-wind-speed.html?m=1
A VESSEL FROM POSITION LAT 45DEG44MINS.N,LONG.150DEG45MINS.W COURSE 290DEG.TRAVEL A 5DAYS IN SPEED OF 18KNOTS. WHAT IS THE ARRIVAL POSITION.
ReplyDeleteYou are in long 126° 35'15"E at LZT 0400H December 11, 2018. What is the LZT and date in long 078° 34'W?
ReplyDeleteAt 0600 zone time, on 22 October you depart Manila, Lat 14°46.0'N Long 118°0'W, with an estimated speed of advance at 20.2 knots. The distance is 6,385.9NM. What is the estimated zone time of arrival at Los Angeles?
ReplyDeletea ship departed south hampton, england with a sailing draft 9.679 meters fwd and 10.065 meters aft. Before arriving Suez Canal for transit, the Chief Officer discharge ballast water about 150 tons distance 80 meters aft of amidship. Find her New Draft by computation.
ReplyDeleteGiven:
DeleteCF = 2m aft of amidship
MTC = 397
TPC = 36.72
LOA = 200m
A survey vessel in position 45° 30’ N, 015° 20’ W steamed north for 48 miles and then steamed east for 60 miles. Find her arrival position?
ReplyDelete1. A ship of 5000 tons displacement has KG 4.2 m and KM 4.5 m, and is listed 5 degrees to port. Assuming that the KM remains constant, find the final list if 80 tons of bunkers are loaded in No. 2 starboard tank whose center of gravity is 1 meter above the keel and 4 meters out from the centerline.
ReplyDeleteParallel Sailing
ReplyDelete1. A vessel sails by parallel sailing from S 18° 22.0’, W 10° 18.0’ to S 18° 22.0’ , W 043° 41.0’
a. Find the True Course
b. Find the distance
A composite great circle track montevido (Lat. 34°55'S Long. 56°10'W) to cape town (Lat.33°55'S Long. 18°25'E) is required with a limiting Latitude of 38°S.Find the total distance to steam the initial course.
ReplyDeleteA landmark is 36.5 meters high and the light has a nominal range of 18 n.m.. Your height of eye is 12.75 meters. If the visibility is 11.2 n.m. when the light becomes visible, approximately how far off the light you will be?
ReplyDeleteplease help me solve this question.
ReplyDeleteyour ship is 30 meters long with 5 meters beam. find the GM if the rolling period is 6 seconds?
A lighthouse bears 2700 T, ship’s head is 0950 T. What is the relative bearing of the lighthouse?
ReplyDeleteOn March 30, 1982 at 1639hrs the Chief Officer asked the Cadet what is Sunset and given position in GPS Latitude 15 degrees 20 minutes North, Longitude 041 degrees 41 minutes East. And what is the Sunrise on March 31 and the given speed was 12 knots and Longitude 041 degrees 41 minutes East.
ReplyDeleteA vessel's light displacement is 2,875 tons. She loaded 390 tons at tons at 2.5 meters. What was the light kG if her KG was 5.20 meters
ReplyDeleteA vessel displacement is 78762 tons has a TPC of 65.3 tons and MTC of 989.4 kilotons. Draft forward reads 12.50 m and draft aft 13.50 m. The trim table for 100 KT shows F.P. (-0.25) and A.P. (3.27) Calculate the new draft fore and aft if you are to discharge 800 tons from cargo hold No.6.
ReplyDeleteIs there already a solution for this one? Thank you for answering
Deletehow to calculate the relative wind direction?is Relative wind direction and true wind direction opposite?
ReplyDeleteA vessel of 11,000 tons displacement and has a cargo on board of 6,000 tons. The existing KG is 28 feet. Solve for the new KG after the 5,000 tons of cargo is shifted up to 10 feet.
ReplyDeleteThe dry bulb thermometer is 80°F and the wet bulb temperature is 74°F, what is the relative humidity?
ReplyDeleteA ship sails from Lat 14°16.2’S with a course of 048° (T) with a speed of 11.6 knots. What is her latitude after 45 minutes?
ReplyDeleteSolve the problem below using Great Circle Sailing
Initial Position, A: (11° 14’ N, 125° 03’ E)
Final Position, B: (08° 01’ S, 079° 34’ W)
Solve the problem below using Great Circle Sailing
ReplyDeleteInitial Position, A: (11° 14’ N, 125° 03’ E)
Final Position, B: (08° 01’ S, 079° 34’ W)
A ship sails from Lat 14°16.2’S with a course of 048° (T) with a speed of 11.6 knots. What is her latitude after 45 minutes?
ReplyDeleteAt about 1300/26th feb 2014 vessel sailed from port saranggani ls. Along Lat/05-24N, long /125- 23E and bound for guam is Along lat/13-28N, Long/144-36E at speed of 12.8 knotsm that the vessel upon arival at guam.
ReplyDeleteVessel sched to berth upon her arival then after 10 days exactly including discharging, vessel sched to depart Guam w/same Lat/Long of above, heading towards hayapura port, papua new guinea along Lat/02-305, long/140-05E.
A. Draw a triangle w/ illustration showing DLAT,DLO,DMP,DEP and Distance.
B. DLAT and DLO between sarangani ls and Guam ls.
C. Course and Distance
D. Steaming time between sarangani ls and guam ls.
E. ETA at Guam
F. ETD at Guam using above 10days stay at Guam including discharging.
G. DLAT and DLO between Guam and Jayaputra port.
H. course and distance.
I. Steeming time between Guam and Jayaputra port, new guinea.
J. ETA jayaputra port using a speed of knots.
referring to the light list a light has a nominal range of 14 miles and 12.8 meters high. if the visibility is 6 nautical miles and your height of eye is 6 meters, at what approximate distance will you sight the light? Formula
ReplyDeleteYour vessel has a loaded draft of 11.5m. You are due to sail on the am from the Port of Weipa on 10th of June 2006. What is the earliest time that you can sail to clear a shoal 12.6 in with minimum 0.5m clearance?
ReplyDeleteTwo box-shaped vessels are each 80 m long, 6 m deep, float at 4 m draft and have KG 3 m. Compare their initial Metacentric Heights if one has 10 m beam and the other has 12 m beam.
ReplyDelete(be sure to look at the examples from the book "ship stability for masters and mates" chapter 12.)
On November 30 the 1430 QR longitude of a ship is 51° 32.4' W. Ten hours later the DR long is 53 07.2' W.
ReplyDeleteRequired: ZT and date of arrival at the second longitude
A rectangular tank (3m 1.2m 0.6 m) has no lid and is floating in fresh
ReplyDeletewater at a draft of 15 cm. Calculate the minimum amount of fresh water
which must be poured into the tank to sink it.
A ship is inclined by an external force to an angle of heel 3°.if the displacement of the ship is 5600 toññes,km is 14.8m and kG is 14.2m.calculate MSS.
ReplyDeleteA product tanker 120m long,15m beam,and 10m depth is floating on an even keel at a draught of 5.50m,block coefficient 0.8 in salt water.fine the cargo to discharge so that the ship will float at the same draft in fresh water
On April 09, 2005 at 2000H UTC. The duty officer took the bearing of the sun which is 277° at position 26°45'N 030°00'W. Having the gyro heading of 030° magnetic heading of 040 and the variation of that position is 1.34°W. What would be the gyrocompass error and the deviation?
ReplyDeleteCan anyone help me to solve this?
This is the sample problem. A landmark is 36.5 meters high and the light has a nominal range of 18 Nautical miles.. Your height of eye is 12.75 meters. If the visibility is 11.2 Nautical miles. when the light becomes visible, approximately how far off the light will be?
ReplyDeletewrite the formula for dynamic stability.a ship of 12000 t displacement has an initial metacentric height of 2m .what is the dynamic stability when the ship is heeled 15 degree
ReplyDelete