§ 06 – 08

 

TIME AND ANGLE MEASUREMENT

20. How is sidereal time measured and why is it important to the navigator?

Sidereal Time is time measured against the fixed stars – a sidereal day being the interval between two successive transits of a star across a given meridian. For convenience, the transit of the First Point of Aries, rather than any particular star, across the upper meridian of Greenwich is used. Sidereal time is important to the navigator because its angular measurement is needed to find the Local Hour Angle (LHA) of a star.

21. Why is it necessary to employ the imaginary body called the Mean Sun?

The Mean Sun is an imaginary body conceived to move at a uniform rate along the celestial horizon taking the same time to complete one revolution about the celestial sphere as does the True Sun in the Ecliptic. Such a body is necessary because the True Sun does not appear to move at a constant rate in the Ecliptic. As the Sun does not lie at the exact centre of the Earth’s orbit, the Earth’s speed is faster when nearest the Sun (Perihelion, in January) and slower when furthest away (Aphelion, in July).

22. Define Local Mean Time (LMT) and Greenwich Mean Time (GMT).

• Local Mean Time is the angle expressed in time contained between the observer’s lower celestial meridian and the meridian of the Mean Sun measured westwards at the centre of the Earth in the plane of the celestial equator.
• Greenwich Mean Time is the angle expressed in time contained between the Greenwich lower celestial meridian and the meridian of the Mean Sun measured westwards at the centre of the Earth in the plane of the celestial equator.

23. What is the difference between Standard Time and Zone Time?

Standard Time is the time kept in various countries and areas throughout the world related to a central meridian of the particular country and a convenient number of hours or half hours different to Greenwich time, whereas Zone Time is time kept in an area lying 7½° of longitude either side of a meridian which is an integral number of hours different to Greenwich Mean Time.

24. Why is it desirable for a chronometer or deck watch to have a constant rate and what precautions can a navigator take to achieve this?

It is desirable for a chronometer to have a constant rate so that its error can be predicted and used with confidence. To achieve this, the gimballed chronometer (or equivalent timepiece) should be stowed in a dust proof locker insulated to maintain as steady a temperature as possible and from the influence of magnetic fields.

25. The adjacent figures show the true and reflected images of a star as seen in a sextant, when the micrometer drum is turned the images move as indicated by the arrows in fig. B. What error, if any, is present here, and what could be done to correct it?

If the true and reflected images of the star move as indicated by the arrows then Side Error is present in the sextant. To correct it, the sextant should be set to zero and the screw on the centre line at the back of the Horizon Glass turned until the two images of the star are coincident. When the micrometer drum is then turned the images will move as shown here.

26. Explain how Index error may be determined by observing the Sun’s diameter.

With the sextant held vertically observe the Sun’s diameter first on the arc, and then off the arc, and divide the difference between the two readings by two, the Index Error being named according to the greater of the two readings. If the Sun is close to the horizon these observations should be made with the sextant held horizontally to avoid any distortion caused by refraction. A check on the accuracy of the observations can be made by dividing the sum of the two readings by four and comparing this with the Sun’s semi-diameter for the day given in the Nautical Almanac.

27. If the Index Error is reasonably small, why is it preferable not to adjust the sextant for it, but simply allow for it when measuring aItitudes?

It is better to leave a small Index Error uncorrected to avoid ‘tormenting’ the sextant which may cause the adjusting screw to wear slack.

28. The adjacent figures show the true and reflected images of the sea horizon when viewed in a sextant held vertically. In fig. A the sextant reads 0° 00′ 00″ and in fig. B it reads 0° 01′ 24″. What does this indicate, and what should be done about it?

The figures illustrate an Index Error of 1’24” on the arc, which would be written IE — 1’24” to indicate that the error should be subtracted from all angles measured with the sextant. In conformity with the answer given to Q8 above, nothing should be done about an error of such small magnitude other than to allow for it when measuring angles. If the sextant has not been adjusted for some considerable time then the Index Error may be removed by turning the off—centre screw at the back of the Horizon Glass.

29. What is the reading on each of the following micrometer sextants?

 

The sextant readings are:

31° 55′ 00″

3° 17′ 50″ off the arc

 

 

30. Without using tables, convert:  (a) 169° 52′ into time (b) 6h 36m 25s into arc

31. Assuming the chronometer to be correct for GMT and the vessel’s clock to be set to mean time for the meridian she is on, what is the vessel’s longitude if the chronometer shows 10h 20m and the clock 13h 35m?

The vessel Longitude must be 48º 45’ E.

32. If the LMT of an observation in Longitude 107°26’E was 15d 5h 44m 165, what was the GMT of this observation?

First convert the longitude into time:

33. If a yacht is in Longitude 132°E and the ZT is 6d 13h 00m, what is the GMT at the same instant?

Longitude 132º E lies between 127½º E and 142½º E, the boundaries of Zone -9

34. In what Time Zone does each of the following Longitudes lie:—

05°32’E          (b) 68°49’W       (c) 100°E (d)    172°29′.5W    (e) 176°27’E

          Zone 0           Zone +5             Zone -7            +11                  -12 (Unless amended by Date Line)

35. On a small vessel with a chronometer (C) and a deck watch (D), the chronometer was gaining 1.5 seconds daily and the deck watch was gaining 3.5s daily. If the chronometer is 3m 2s fast of the deck watch, what will be the error of the deck watch on the chronometer at the end of a further 40 days?

36. On Jun 25th 199X in Long 42° 10′ W at 09:35 approximate time an observation gave the Sun’s LHA to be 322°30′.2 when the chronometer indicated 12h 18m 40s. Find the chronometer error.

37. At about 19:00 on Sep 17th in Long 162°00′ W a chronometer read 05h 40m 10s. On May 17th  this chronometer was 4m 10s fast of GMT and on Jul 03rd it was 2m 32s fast of GMT (all comparisons at 12:00 GMT). Find the correct GMT on Sep 17th .

38. Find the GMT on a vessel where an observation on Sep 03rd gave the time by chronometer as 3d 0h 8m 44s. which on Aug 01st was 2m 11s fast at 12:00 GMT and was losing 3.3s daily.

39. Find the GMT on the Mar 17th , 1992, when the time by chronometer was 17d 16h 10m 21s, which was 2m 21s fast at 12:00 GMT on Dec 20th previously and 1m 09s fast at 12:00 GMT on Feb 03rd , 1992.

Required GMT = 17d 16h 10m 21s

40. Without using tables convert: – 169° 52.0 into time.

41. 6h. 36m. 25s. into arc.

∴ 6h 36m 25s into ARC = 99° 06’ 15”

42. Assuming the chronometer to be correct for G.M.T. and the vessel’s clock. to be set to mean time for the meridian she is on. What is the vessel’s Longitude if the chronometer shows 10h. 20m. and the clock.13h. 35m?

Chronometer correct to G.M.T  = 10h 20m

Vessel clock set to L.M.T.   = 13h 35m

∴   G.M.T. is less than L.M.T. Longitude is therefore East

43. If the L.M.T. of an observation in Longitude 107° 26.0 E. was 15d. 5h. 44m.16s. what was the G.M.T. of this observation?

∴            107° 26’ into time = x 4 ÷ 60 L.M.T.  = 24h 44m 76s

44. It a ship is in Longitude 132° E. and the ZT is 6d.13h. 00m. what is the G.M.T. at the same instant?

Longitude 132° E                = Zone (-9) = + on G.M.T.

Zone Time                           = 6d. 13h. 00m. 

Zone Time                           = 6d. 13h. 00m.

Zone Description                            = (-)   09h.

Greenwich Date + G.M.T.  = 6d. 04h. 00m. G.M.T.     What would the L.M.T. for this question be?

45. Find the GMT. on a vessel where an observation on Sep 03rd gave the time by chronometer as 3d. 0h. 8m. 14s. which on Aug 01st at 12:00 GMT. and was losing 3.3 seconds daily.

To Find Accumulated Rate:

∴ G.M.T. Sep 03rd = 3d. 0h. 08m. 20s.

46. On Oct 30th on a vessel when the time by chronometer was 29d. 22h. 24m. 36s. and which was 2m.18s. slow at 12:00 GMT. on Aug 08th , and 1m. 32s. fast at 12:00 GMT. on Sep 27th . Find the GMT. of observation.

Oct 30th 29d. 22h. 24m. 36s.

12:00 Hrs GMT. Aug 08th  02m. 18s. Slow  ɿ  Number of Days between Errors = 50 Days

12:00 Hrs GMT. Sep 27th  01m. 32s. Fast    ȷ

∴  G.M.T. Oct 30th = 29d. 22h. 20m. 35s.

47. Jul 19th 20d. 4h. 21m. 36s. 12:00 Hrs GMT. Jun 14th. 1m. 35s. SLOW Gaining 2.5s. daily.

To Find Accumulated Rate:

∴ GMT Jul 19th = 20d. 4h. 21m. 42.

48. Jan 05th 5d. 16h. 13m. 55s. 12:00 Hrs GMT. Dec 20th 0m. 58s. FAST Losing 1.5s. daily.

To Find Accumulated Rate:

49. Nov 22nd 22d. 11h. 45m. 36s. 12:00 Hrs GMT. Jul 25th 4m. 36s. SLOW 12:00 Hrs GMT. Sep 08th 2m. 03s. Slow.

∴   GMT Nov 22nd = 22d 11h. 43m. 24s.

50. Mar 08th 8d. 17h. 44m. 36s. 12:00 Hrs GMT. Dec 22nd 3m. 46s. FAST. 12:00 Hrs GMT. Jan 29th 2m. 49s. Fast. Error = 0m 57s. Slow.

Number of Days Dec 22nd – Jan 29th = 38 Days

∴    GMT Mar 08th = 8d. 17h. 42m. 44s.