TAB 4 Celestial Navigation - Chapter 18 - Time.pdf - N. Bowditch - towotowo

CHAPTER 18

TIME

TIME IN NAVIGATION

1800. Solar Time

moved to point C in its orbit. Thus, during the course of a day

the sun appears to move eastward with respect to the stars.

The apparent positions of the stars are commonly reck-

oned with reference to an imaginary point called the vernal

equinox , the intersection of the celestial equator and the

ecliptic. The period of the earth’s rotation measured with

respect to the vernal equinox is called a sidereal day . The

period with respect to the sun is called an apparent solar

day .

When measuring time by the earth’s rotation, using the

actual position of the sun results in apparent solar time .

Use of the apparent sun as a time reference results in

time of non-constant rate for at least three reasons. First, rev-

olution of the earth in its orbit is not constant. Second, time

is measured along the celestial equator and the path of the

real sun is not along the celestial equator. Rather, its path is

along the ecliptic, which is tilted at an angle of 23

The earth’s rotation on its axis causes the sun and other

celestial bodies to appear to move across the sky from east

to west each day. If a person located on the earth’s equator

measured the time interval between two successive transits

overhead of a very distant star, he would be measuring the

period of the earth’s rotation. If he then made a similar mea-

surement of the sun, the resulting time would be about 4

minutes longer. This is due to the earth’s motion around the

sun, which continuously changes the apparent place of the

sun among the stars. Thus, during the course of a day the

sun appears to move a little to the east among the stars so

that the earth must rotate on its axis through more than 360

in order to bring the sun overhead again.

See Figure 1800. If the sun is on the observer’s meridian

when the earth is at point A in its orbit around the sun, it will

not be on the observer’s meridian after the earth has rotated

through 360

27' with

respect to the celestial equator. Third, rotation of the earth

on its axis is not constant.

To obtain a constant rate of time, the apparent sun is re-

placed by a fictitious mean sun . This mean sun moves

eastward along the celestial equator at a uniform speed equal

because the earth will have moved along its or-

bit to point B. Before the sun is again on the observer’s

meridian, the earth must turn still more on its axis. The sun

will be on the observer’s meridian again when the earth has

Figure 1800. Apparent eastward movement of the sun with respect to the stars.

287

288

TIME

to the average speed of the apparent sun along the ecliptic.

This mean sun, therefore, provides a uniform measure of

time which approximates the average apparent time. The

speed of the mean sun along the celestial equator is 15

per

Example 2: See Figure 1801. Determine the time of the up-

per meridian passage of the sun on April 16, 1995.

Solution: From Figure 1801, upper meridian passage

of the sun on April 16, 1995, is given as 1200. The dividing

line between the values for upper and lower meridian pas-

sage on April 16th indicates that the sign of the equation of

time changes between lower meridian passage and upper

meridian passage on this date; the question, therefore, be-

comes: does it become positive or negative? Note that on

April 18, 1995, upper meridian passage is given as 1159,

indicating that on April 18, 1995, the equation of time is

positive. All values for the equation of time on the same side

of the dividing line as April 18th are positive. Therefore, the

equation of time for upper meridian passage of the sun on

April 16, 1995 is (+) 00 m 05 s . Upper meridian passage,

therefore, takes place at 11 h 59 m 55 s .

hour of mean solar time.

1801. Equation Of Time

Mean solar time ,or mean time as it is commonly

called, is sometimes ahead of and sometimes behind appar-

ent solar time. This difference, which never exceeds about

16.4 minutes, is called the equation of time .

The navigator most often deals with the equation of time

when determining the time of upper meridian passage of the

sun. The sun transits the observer’s upper meridian at local ap-

parent noon . Were it not for the difference in rate between the

mean and apparent sun, the sun would be on the observer’s me-

ridian when the mean sun indicated 1200 local time. The

apparent solar time of upper meridian passage, however, is off-

set from exactly 1200 mean solar time. This time difference, the

equation of time at meridian transit, is listed on the right hand

daily pages of the Nautical Almanac .

The sign of the equation of time is positive if the time

of sun’s meridian passage is earlier than 1200 and negative

if later than 1200. Therefore: Apparent Time = Mean Time

– (equation of time).

SUN

MOON

Day

Eqn. of Time

Mer.

Mer. Pass.

00 h

12 h

Pass.

Upper Lower Age Phase

m d

00 02 00 05 12 00 00 26 12 55 16

00 13 00 20 12 00 01 25 13 54 17

00 27 00 33 11 59 02 25 14 55 18

Figure 1801. The equation of time for April 16, 17, 18, 1995.

Example 1: Determine the time of the sun’s meridian

passage (Local Apparent Noon) on June 16, 1994.

Solution: See Figure 2007 in Chapter 20, the Nautical

Almanac’s right hand daily page for June 16, 1994. The

equation of time is listed in the bottom right hand corner of

the page. There are two ways to solve the problem, depend-

ing on the accuracy required for the value of meridian

passage. The time of the sun at meridian passage is given to

the nearest minute in the “Mer. Pass.”column. For June

16, 1994, this value is 1201.

To determine the exact time of meridian passage, use

the value given for the equation of time. This value is listed

immediately to the left of the “Mer. Pass.” column on the

daily pages. For June 16, 1994, the value is given as 00 m 37 s .

Use the “12 h ” column because the problem asked for merid-

ian passage at LAN. The value of meridian passage from the

“Mer. Pass.” column indicates that meridian passage oc-

curs after 1200; therefore, add the 37 second correction to

1200 to obtain the exact time of meridian passage. The exact

time of meridian passage for June 16, 1994, is 12 h 00 m 37 s .

To calculate latitude and longitude at LAN, the navigator

seldom requires the time of meridian passage to accuracies

greater than one minute. Therefore, use the time listed under

the “Mer. Pass.” column to estimate LAN unless extraordinary

accuracy is required.

1802. Fundamental Systems Of Time

The first fundamental system of time is Ephemeris

Time (ET) . Ephemeris Time is used by astronomers in cal-

culating the fundamental ephemerides of the sun, moon,

and planets. It is not used by navigators.

The fundamental system of time of most interest to

navigators is Universal Time (UT) . UT is the mean solar

time on the Greenwich meridian, reckoned in days of 24

mean solar hours beginning with 0 h at midnight. Universal

Time, in principle, is determined by the average rate of the

apparent daily motion of the sun relative to the meridian of

Greenwich; but in practice the numerical measure of Uni-

versal Time at any instant is computed from sidereal time.

Universal Time is the standard in the application of astron-

omy to navigation. Observations of Universal Times are

made by observing the times of transit of stars.

The Universal Time determined directly from astro-

nomical observations is denoted UT0 . Since the earth’s

rotation is nonuniform, corrections must be applied to UT0

to obtain a more uniform time. This more uniform time is

obtained by correcting for two known periodic motions.

One motion, the motion of the geographic poles, is the

result of the axis of rotation continuously moving with re-

The equation of time’s maximum value approaches

16 m 22 s in November.

If the Almanac lists the time of meridian passage as

1200, proceed as follows. Examine the equations of time list-

ed in the Almanac to find the dividing line marking where the

equation of time changes between positive and negative val-

ues. Examine the trend of the values near this dividing line to

determine the correct sign for the equation of time.

TIME

289

spect to the earth’s crust. The corrections for this motion are

quite small (

4 m

=60'

15 milliseconds for Washington, D.C.). On

applying the correction to UT0, the result is UT1 , which is

the same as Greenwich mean time (GMT) used in celestial

navigation.

The second known periodic motion is the variation in

the earth’s speed of rotation due to winds, tides, and other

phenomena. As a consequence, the earth suffers an annual

variation in its speed of rotation, of about

60 s

=1 m

= 15'

4 s

= 1'

= 60"

1 s

= 15"

= 0.25'

30 milliseconds.

When UT1 is corrected for the mean seasonal variations in

the earth’s rate of rotation, the result is UT2 .

Although UT2 was at one time believed to be a uni-

form time system, it was later determined that there are

variations in the earth’s rate of rotation, possibly caused by

random accumulations of matter in the convection core of

the earth. Such accumulations would change the earth’s

moment of inertia and thus its rate of rotation.

The third fundamental system of time, Atomic Time

(AT) , is based on transitions in the atom. The basic princi-

ple of the atomic clock is that electromagnetic waves of a

particular frequency are emitted when an atomic transition

occurs. The frequency of the cesium beam atomic clock is

9,192,631,770 cycles per second of Ephemeris Time.

The advent of atomic clocks having accuracies better

than 1 part in 10 -13 led in 1961 to the coordination of time

and frequency emissions of the U. S. Naval Observatory and

the Royal Greenwich Observatory. The master oscillators

controlling the signals were calibrated in terms of the cesi-

um standard, and corrections determined at the U. S. Naval

Observatory and the Royal Greenwich Observatory were

made simultaneously at all transmitting stations. The result

is Coordinated Universal Time (UTC) .

Therefore any time interval can be expressed as an

equivalent amount of rotation, and vice versa. Interconver-

sion of these units can be made by the relationships

indicated above.

To convert time to arc:

1. Multiply the hours by 15 to obtain degrees of arc.

2. Divide the minutes of time by four to obtain

degrees.

3. Multiply the remainder of step 2 by 15 to obtain

minutes of arc.

4. Divide the seconds of time by four to obtain min-

utes of arc

5. Multiply the remainder by 15 to obtain seconds of arc.

6. Add the resulting degrees, minutes, and seconds.

Example 1: Convert 14 h 21 m 39 s to arc.

Solution:

(1) 14 h

= 210

00' 00"

(2) 21 m

= 005

00' 00" (remainder 1)

(3) 1

= 000

15' 00"

(4) 39 s

= 000

09' 00" (remainder 3)

(5) 3

= 000

00' 45"

1803. Time And Arc

(6) 14 h 21 m 39 s

= 215

24' 45"

One day represents one complete rotation of the earth.

Each day is divided into 24 hours of 60 minutes; each

minute has 60 seconds.

Time of day is an indication of the phase of rotation of

the earth. That is, it indicates how much of a day has

elapsed, or what part of a rotation has been completed.

Thus, at zero hours the day begins. One hour later, the earth

has turned through 1/24 of a day, or 1/24 of 360

To convert arc to time:

, or 360

°¸

1. Divide the degrees by 15 to obtain hours.

2. Multiply the remainder from step 1 by four to ob-

tain minutes of time.

3. Divide the minutes of arc by 15 to obtain minutes

of time.

4. Multiply the remainder from step 3 by four to ob-

tain seconds of time.

5. Divide the seconds of arc by 15 to obtain seconds

of time.

6. Add the resulting hours, minutes, and seconds.

Smaller intervals can also be stated in angular units;

since 1 hour or 60 minutes is equivalent to 15

, 1 minute of

time is equivalent to 15

°¸

60 = 0.25

= 15', and 1 second

of time is equivalent to 15'

60 = 0.25' = 15".

Example 2: Convert 215

24' 45" to time units.

Summarizing in table form:

Solution:

(1) 215

°¸

= 14 h 00 m 00 s

remainder 5

Time

Arc

(2) 5

= 00 h 20 m 00 s

1 d

=24 h

=360

(3) 24'

= 00 h 01 m 00 s

remainder 9

(4) 9

= 00 h 00 m 36 s

60 m

=1 h

=15

24 = 15

290

TIME

(5) 45"

= 00 h 00 m 03 s

mediately to the east of the line. When solving problems,

convert local time to Greenwich time and then convert this to

local time on the opposite side of the date line.

(6) 215

24' 45" = 14 h 21 m 39 s

Solutions can also be made using arc to time conversion

tables in the almanacs. In the Nautical Almanac , the table

given near the back of the volume is in two parts, permitting

separate entries with degrees, minutes, and quarter minutes

of arc. This table is arranged in this manner because the nav-

igator converts arc to time more often than the reverse.

1806. Zone Time

is usually used as

the time meridian or zone meridian . Thus, within a time

zone extending 7.5' on each side of the time meridian the

time is the same, and time in consecutive zones differs by

exactly one hour. The time is changed as convenient, usu-

ally at a whole hour, when crossing the boundary between

zones. Each time zone is identified by the number of times

the longitude of its zone meridian is divisible by 15

18'22" to time units, using the

Nautical Almanac arc to time conversion table.

Solution:

, posi-

tive in west longitude and negative in east longitude . This

number and its sign, called the zone description (ZD) ,is

the number of whole hours that are added to or subtracted

from the zone time to obtain Greenwich mean time (GMT).

The mean sun is the celestial reference point for zone time.

See Figure 1806.

Converting ZT to GMT, a positive ZT is added and a

negative one subtracted; converting GMT to ZT, a positive

ZD is subtracted, and a negative one added.

Convert the 22" to the nearest quarter minute of arc for

solution to the nearest second of time. Interpolate if more

precise results are required.

334

00.00 m

= h 16 m 00 s

000

18.25 m

= 00 h 01 m 13 s

334

18' 22"

= 22 h 17 m 13 s

1804. Time And Longitude

Example: The GMT is 15 h 27 m 09 s .

Suppose a celestial reference point were directly over

a certain point on the earth. An hour later the earth would

have turned through 15

Required: (1) ZT at long. 156

24.4' W.

, and the celestial reference would

be directly over a meridian 15

(2) ZT at long. 039

04.8' E.

farther west. Any difference

of longitude between two points is a measure of the angle

through which the earth must rotate to separate them.

Therefore, places east of an observer have later time, and

those west have earlier time, and the difference is exactly

equal to the difference in longitude, expressed in time units.

The difference in time between two places is equal to the

difference of longitude between their meridians, expressed

in time units instead of arc.

Solutions:

(1)

GMT

15 h 27 m 09 s

+10 h (rev.)

05 h 27 m 09 s

(2)

GMT

15 h 27 m 09 s

–03 h (rev.)

1805. The Date Line

18 h 27 m 09 s

Since time is later toward the east and earlier toward the

west of an observer, time at the lower branch of one’s merid-

ian is 12 hours earlier or later depending upon the direction

of reckoning. A traveler making a trip around the world gains

or loses an entire day. To prevent the date from being in error,

and to provide a starting place for each day, a date line is

fixed by international agreement. This line coincides with the

180th meridian over most of its length. In crossing this line,

the date is altered by one day. If a person is traveling east-

ward from east longitude to west longitude, time is becoming

later, and when the date line is crossed the date becomes 1

day earlier. At any moment the date immediately to the west

of the date line (east longitude) is 1 day later than the date im-

1807. Chronometer Time

Chronometer time (C) is time indicated by a chro-

nometer. Since a chronometer is set approximately to GMT

and not reset until it is overhauled and cleaned about every

3 years, there is nearly always a chronometer error (CE) ,

either fast (F) or slow (S). The change in chronometer error

in 24 hours is called chronometer rate ,or daily rate , and

designated gaining or losing. With a consistent rate of 1 s per

day for three years, the chronometer error would be approx-

imately 18 m . Since chronometer error is subject to change,

it should be determined from time to time, preferably daily

at sea. Chronometer error is found by radio time signal, by

At sea, as well as ashore, watches and clocks are nor-

mally set to some form of zone time (ZT) . At sea the

nearest meridian exactly divisible by 15

Example 3: Convert 334

Figure 1806. Time Zone Chart.

TAB 4 Celestial Navigation - Chapter 18 - Time.pdf

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