Zone time

ZONE TIME Zone Time What does it mean? For our purposes, ALL time is based on the GREENWICH Meridian which is 00, where time began as we discovered when we were discussing the “chronometer” Zone Time As a result, each Meridian of Longitude has it’s own Local Mean Time (LMT (LMT)) eg eg while Mumbai has a local time of 12:30 the corresponding time in London would be 07:00 (Mumbai being 5.5hrs ahead of GMT in time.) Zone Time So, as a ship travels eastwards from London to Mumbai the ship’s clocks (NOT the chronometer please remember!) have to be advanced so that when the ship arrives in Mumbai it is keeping the same time as Mumbai, in other words the Local Time or LMT LMT Greenwich Obsevatory Where “time” for the Mariner first began with the invention of the Chronometer Zone Time In the same way for example, if the time in London is 12:00 the corresponding Local Time (or LMT LMT)) in New York is 07:00 (New York being 5hrs “behind” GMT) So, as a ship travels westward from London to New York the ship’s clocks have to be retarded so that when the ship arrives in New York it is keeping the same time as New York i.e LMT LMT Zone Time In the shipping industry we cannot get away from this continuous changing of the “Ship’s Time” (ST) as the ship crosses Meridians while travelling eastward or westward Zone Time In order to simplify it to a certain extent and also to aid navigation, the world is divided into 24 ‘Time Zones’, each of these Zones extending over 150 of Longitude and representing hour of time Zone Time The time being kept in each of these Zones being the LMT of the central Meridian in each Zone So, if we take the Greenwich Meridian (00) as the base line, with 7.50 either side we have our first Zone which is Greenwich Mean Time (GMT or UTC) UTC) as follows:follows:- Zone Time N POLE 150 7.50 E 7.50 W 00 Consider the following:- Two ships keeping Zone Time (ZT) C 1800 NP A 00 G B Two ships, A & B, keeping ZT and same speed leave ‘G’ for ‘C’ on Sat.1st @ 1100Z and each take 28 days For ‘A’ :Dep 1d 11h 00m Z + 28d (steaming time) Arr 29d 11h 00m Z (Travels west so clocks go back, therefore:-) ZT (-) 12h 00m Arr 28d 23h 00m ZT Similarly For ship ‘B’ C 1800 NP A 00 G B Two ships, A & B, keeping ZT and same speed leave ‘G’ for ‘C’ on Sat.1st @ 1100Z and each take 28 days Dep 1d 11h 00m Z + 28d (steaming time) Arr 29d 11h 00m Z (Travels east so clocks go forward, therefore:-) ZT (+) 12h 00m Arr 29d 23h 00m ZT The International Date Line • Ship A arrived @ • 28d 23h 00m • If it crosses the IDL we must +/- one day • 28d 23h 00m (+) • 01d • 29d 23h 00m • Ship B arrived @ • 29d 23h 00m • If it crosses the IDL we must +/- one day • 29d 23h 00m (-) • 01d • 28d 23h 00m Zone Time Example A vessel sails from San Francisco (USA) at 1300 ST on the 16th November 1979 bound for Yokohama, Japan Find the ETA, GMT and Standard Time (ST) at Yokohama The total distance is 7642’ and the average speed is 15.4kn Example Distance (D) = Speed (S) x Time (T), T = D/S which is = 7642/15.4 which is = 496.233hrs Therefore voyage time = 20d 16h 14m San Francisco (SF) Going W to E therefore we must add day? Yokohama (Y) IDL (International Date Line) Dep SF: ZT: Dep SF: Voy Time: Arr Y: Arr Y: 16d 13h 8h 16d 21h 20d 16h 37d 13h 07d 13h 00m ST (LMT) 00m (+) 00m Z (Nov) 14m (+) 14m Z 14m Z Arr Y: 07d 13h 14m Z ZT : 09h 00m Arr Y: 07d 22h 14m LMT = ST Example A vessel sails from San Diego (USA) at 1500 ST on the 6th July 2006 bound for Brisbane, Australia Find the ETA, GMT an Standard Time (ST) at Brisbane The total distance is 8625’ and the average speed is 14.2kn Example Going W to E therefore we must add 1day? ►Steaming Time = D/S ►= 8625/14.2 ►= 607.394 hrs ►= 25d 07h 24m Dep SD: ►ZT : ►Dep SD: ►Voy Time: ►Arr Br : ►Arr Br : ►ZT : ►Arr Br: ► Brisbane (Br) IDL San Diego (SD) 06d 15h 00m ST ( LMT) 8h 00m (+) 06d 23h 00m Z 25d 07h 24m (+) 32d 06h 24m Z 01d 06h 24m Z 10h 00m + 01d 16h 24m LMT or ST Going W to E therefore we must add 1day? ►Steaming Time = D/S ►= 8625/14.2 ►= 607.394 hrs ►= 25d 07h 24m Dep SD ►Voy Time ►Arr Br ►Clock change ►Arr Br ►Plus 1day ►Arr Br ►ZT ►Arr Br ► Brisbane (Br) IDL San Diego (SD) Alternatively, using Ship’s Time :06d 15h 00m ST ( LMT) :25d 07h 24m (+) :31d 22h 24m ST : 6h ((-) :31d 16h 24m ST :01d (x’ing IDL) :01d 06h 24m Z :10h 00m + :01d 16h 24m LMT or ST To Find the Correct U.T For most navigation questions it is required to know a GMT The question may sometimes provide this However, the most common presentations are as follows:follows:- To Find the Correct U.T Date & time at the ship AND a chronometer time OR! AM or PM twighlight To Find the Correct U.T In such cases it is necessary to find an approximate GMT to decide upon the date and, because the chronometer is not a 24hour clock whether it is reading AM or PM This is particularly important when large values for longitude are encoutered To Find the Correct U.T Example 1:1:Ship’s Time is 1840, January 15th in DR position 150 00’N ; 1700 00’W The chronometer read 5h 59m 10s which was 2m 20s slow on GMT Find the correct GMT To Find the Correct U.T LMT: 15d Long W (in time) ~GMT: 15d ~GMT: 16d Therefore:-Therefore: 16d Chron: Error: GMT: 16d 18h 11h 30h 06h 40m 20m 00m 00m 00s ? (+) 00s 00s 05h 59m 10s 02m 20s ? 06h 01m 30s (+) To Find the Correct U.T Example 2:2:On the 21st of November 2000, an AM sight of Polaris was taken in DR 450 00’N ; 1500 00’E The chronometer showed 8h 37m 10s which was 1m 8s fast on GMT Find the correct GMT To Find the Correct U.T LMT twilight: 21d 06h Long E (in time) 10h ~GMT: 20d 20h Therefore:-Therefore: Chron: 20d 20h Error: GMT: 20d 20h 34m 00s 00m ? (-) 34m 00s 37m 10s 01m 08s ? (-) 36m 02s [...]... 300 W/E And so on as we will see from the following:following:- Zone Time 1800 -11 -12 +12 +11 +10 -10 -9 +9 -8 +8 -7 N 900 -6 +6 POLE +5 -5 +4 -4 +3 -3 150 +2 -2 -1 0 00 +1 900 +7 Time Zones • Represented on a chart it would look something like this:- Click 2nd time round Zone Time The western Zones are 12 hours behind GMT and: The eastern Zones are 12 hours ahead of GMT Thus, a 24 hour (or 1 day) difference... departs from the 180th Merdian in the region of E Siberia, the Aleutian Islands and some groups of Pacific Islands Zone Time STANDARD TIMES (ST) Zone Time may be modified to suit times kept between boundaries of various countries or states These time are known as Standard Times A list of Standard Times can be found at the back of the Nautical Almanac Consider the following:- Two Aircraft keeping GMT (Z).. .Zone Time Each subsequent Zone, whether travelling East or West, is also divided into 150 sectors and will either be “ahead” or “behind” GMT respectively i.e Zone 0, from 7.50 east to 7.50 west, keeps GMT Zone Time Zones -1 and +1 (7.50 W/E – 22.50 W/E) will be one hour behind or ahead of GMT respectively and keeping the LMT of 150 W/E Zones -2 and +2 (22.50 W/E – 37.50... Both take the same time i.e 10hours • Therefore:• ‘A’ arrives at 1100z + 10hours = 2100Z on 5th • ‘B’ arrives at 1100z + 10hours = 2100Z on 5th Where does that leave us? From this we can see that if an aircraft or a ship continually keeps GMT it is not necessary to worry about Zone Time Unfortunately, the time onboard will be totally out of synch with Local Time or LMT or Standard Time for whatever... halves To account for this difference we need to apply the following:following:- Zone Time When crossing the 1800 meridian from West to East we have to ADD one day When crossing the 1800 meridian from East to West we have to SUBTRACT one day This essentially gives us what we call the International Date Line or IDL IDL Zone Time The IDL mainly follows the 1800 meridian but to avoid division of geographically... Find the ETA, GMT and Standard Time (ST) at Yokohama The total distance is 7642’ and the average speed is 15.4kn Example Distance (D) = Speed (S) x Time (T), T = D/S which is = 7642/15.4 which is = 496.233hrs Therefore voyage time = 20d 16h 14m 1 San Francisco (SF) Going W to E therefore we must add 1 day? Yokohama (Y) IDL (International Date Line) Dep SF: ZT: Dep SF: Voy Time: Arr Y: Arr Y: 16d 13h... + 28d (steaming time) Arr 29d 11h 00m Z (Travels east so clocks go forward, therefore:-) ZT (+) 12h 00m Arr 29d 23h 00m ZT The International Date Line • Ship A arrived @ • 28d 23h 00m • If it crosses the IDL we must +/- one day • 28d 23h 00m (+) • 01d • 29d 23h 00m • Ship B arrived @ • 29d 23h 00m • If it crosses the IDL we must +/- one day • 29d 23h 00m (-) • 01d • 28d 23h 00m Zone Time Example 1 A... Time or LMT or Standard Time for whatever area it happens to be in Consider the following:- Two ships keeping Zone Time (ZT) C 1800 NP A 00 G B Two ships, A & B, keeping ZT and same speed leave ‘G’ for ‘C’ on Sat.1st @ 1100Z and each take 28 days For ‘A’ :Dep 1d 11h 00m Z + 28d (steaming time) Arr 29d 11h 00m Z (Travels west so clocks go back, therefore:-) ZT (-) 12h 00m Arr 28d 23h 00m ZT Similarly... 1500 ST on the 6th July 2006 bound for Brisbane, Australia Find the ETA, GMT an Standard Time (ST) at Brisbane The total distance is 8625’ and the average speed is 14.2kn Example 2 Going W to E therefore we must add 1day? ►Steaming Time = D/S ►= 8625/14.2 ►= 607.394 hrs ►= 25d 07h 24m Dep SD: ►ZT : ►Dep SD: ►Voy Time: ►Arr Br : ►Arr Br : ►ZT : ►Arr Br: ► Brisbane (Br) IDL San Diego (SD) 06d 15h 00m... 24m Z 01d 06h 24m Z 10h 00m + 01d 16h 24m LMT or ST Going W to E therefore we must add 1day? ►Steaming Time = D/S ►= 8625/14.2 ►= 607.394 hrs ►= 25d 07h 24m Dep SD ►Voy Time ►Arr Br ►Clock change ►Arr Br ►Plus 1day ►Arr Br ►ZT ►Arr Br ► Brisbane (Br) IDL San Diego (SD) Alternatively, using Ship’s Time :06d 15h 00m ST ( LMT) :25d 07h 24m (+) :31d 22h 24m ST : 6h ((-) :31d 16h 24m ST :01d (x’ing IDL)